Coordinate Geometry Class 10th Mathematics Tamilnadu Board Solution

Class 10th Mathematics Tamilnadu Board Solution
Exercise 5.1
  1. Find the midpoint of the line segment joining the points (i) (1, - 1) and (-5,…
  2. Find the centroid of the triangle whose vertices are (i) (1,3), (2, 7) and (12,…
  3. The center of a circle is at (-6, 4). If one end of a diameter of the circle is…
  4. If the centroid of a triangle is at (1, 3) and two of its vertices are (-7, 6)…
  5. Using the section formula, show that the points A (1,0), B (5,3), C (2,7) and D…
  6. Find the coordinates of the point which divides the line segment joining (3, 4)…
  7. Find the coordinates of the point which divides the line segment joining (-3, 5)…
  8. Let A (-6, -5) and B (-6, 4) be two points such that a point P on the line AB…
  9. Find the points of trisection of the line segment joining the points A (2, -2)…
  10. Find the points which divide the line segment joining A (-4 ,0) and B (0,6)…
  11. Find the ratio in which the x-axis divides the line segment joining the points…
  12. In what ratio is the line joining the points (-5, 1) and (2, 3) divided by the…
  13. Find the length of the medians of the triangle whose vertices are (1, -1), (0,…
Exercise 5.2
  1. (0, 0), (3, 0) and (0, 2) Find the area of the triangle formed by the points…
  2. (5, 2), (3, -5) and (-5, -1) Find the area of the triangle formed by the points…
  3. (-4, -5), (4, 5) and (-1, -6) Find the area of the triangle formed by the…
  4. Vertices: (0, 0), (4, a), (6, 4) Area (in sq. units): 17 Vertices of the…
  5. Vertices: (a, a), (4, 5), (6, -1) Area (in sq. units): 9 Vertices of the…
  6. Vertices: (a, -3), (3, a), (-1,5) Area (in sq. units): 12 Vertices of the…
  7. (4, 3), (1, 2) and (-2, 1) Determine if the following set of points are…
  8. (-2, -2), (-6, -2) and (-2, 2) Determine if the following set of points are…
  9. (- 3/2 , 3) , (6, -2) and (-3, 4) Determine if the following set of points are…
  10. (k, -1), (2, 1) and (4, 5) In each of the following, find the value of k for…
  11. (2, - 5), (3, - 4) and (9, k) In each of the following, find the value of k for…
  12. (k, k), (2, 3) and (4, - 1) In each of the following, find the value of k for…
  13. (6, 9), (7, 4), (4,2) and (3,7) Find the area of the quadrilateral whose…
  14. (-3, 4), (-5, - 6), (4, - 1) and (1, 2) Find the area of the quadrilateral…
  15. (-4, 5), (0, 7), (5, - 5) and (-4, - 2) Find the area of the quadrilateral…
  16. If the three points (h, 0), (a, b) and (0, k) lie on a straight line, then using…
  17. Find the area of the triangle formed by joining the midpoints of the sides of a…
Exercise 5.3
  1. Find the angle of inclination of the straight line whose slope is (i) 1 (ii) √3…
  2. Find the slope of the straight line whose angle of inclination is (i) 30° (ii)…
  3. Find the slope of the straight line passing through the points (i) (3, -2) and…
  4. Find the angle of inclination of the line passing through the points (i) (1, 2)…
  5. Find the slope of the line which passes through the origin and the midpoint of…
  6. The side AB of a square ABCD is parallel to x-axis. Find the (i) slope of AB…
  7. The side BC of an equilateral 3ABC is parallel to x-axis. Find the slope of AB…
  8. (2, 3), (3, -1) and (4, -5) Using the concept of slope, show that each of the…
  9. (4, 1), (-2, -3) and (-5, -5) Using the concept of slope, show that each of the…
  10. (4, 4), (-2, 6) and (1, 5) Using the concept of slope, show that each of the…
  11. If the points (a, 1), (1, 2) and (0, b + 1) are collinear, then show that 1/a +…
  12. The line joining the points A (-2, 3) and B (a, 5) is parallel to the line…
  13. The line joining the points A (0, 5) and B (4, 2) is perpendicular to the line…
  14. The vertices of ΔABC are A (1, 8), B (-2, 4), C (8, -5). If M and N are the…
  15. A triangle has vertices at (6, 7), (2, -9) and (-4, 1). Find the slopes of its…
  16. The vertices of a ΔABC are A (-5, 7), B (-4, -5) and C (4, 5). Find the slopes…
  17. Using the concept of slope, show that the vertices (1, 2), (-2, 2), (-4, -3)…
  18. Show that the opposite sides of a quadrilateral with vertices A (-2, -4), B (5,…
Exercise 5.4
  1. Write the equations of the straight lines parallel to x- axis which are at a…
  2. Find the equations of the straight lines parallel to the coordinate axes and…
  3. Find the equation of a straight line whose (i) slope is -3 and y-intercept is 4.…
  4. Find the equation of the line intersecting the y- axis at a distance of 3 units…
  5. Find the slope and y-intercept of the line whose equation is (i) y = x + 1 (ii)…
  6. Find the equation of the straight line whose (i) slope is -4 and passing through…
  7. Find the equation of the straight line which passes through the midpoint of the…
  8. Find the equation of the straight line passing through the points (i) (-2, 5)…
  9. Find the equation of the median from the vertex R in a ΔPQR with vertices at…
  10. By using the concept of the equation of the straight line, prove that the given…
  11. Find the equation of the straight line whose x and y-intercepts on the axes are…
  12. Find the x and y intercepts of the straight line (i) 5x + 3y - 15 = 0 (ii) 2x -…
  13. Find the equation of the straight line passing through the point (3, 4) and has…
  14. Find the equation of the straight lines passing through the point (2, 2) and…
  15. Find the equation of the straight line passing through the point (5, -3) and…
  16. Find the equation of the line passing through the point (9, -1) and having its…
  17. A straight line cuts the coordinate axes at A and B. If the midpoint of AB is…
  18. Find the equation of the line passing through (22, -6) and having intercept on…
  19. If A(3, 6) and C(-1, 2) are two vertices of a rhombus ABCD, then find the…
  20. Find the equation of the line whose gradient is 3/2 and which passes through P,…
Exercise 5.5
  1. 3x + 4y - 6 = 0 Find the slope of the straight line
  2. y = 7x + 6 Find the slope of the straight line
  3. 4x = 5y + 3. Find the slope of the straight line
  4. Show that the straight lines x + 2y + 1 = 0 and 3x + 6y + 2 = 0 are parallel.…
  5. Show that the straight lines 3x - 5y + 7 = 0 and 15x + 9y + 4 = 0 are…
  6. If the straight lines y/2 = x-p and ax + 5 = 3y are parallel, then find a.…
  7. Find the value of a if the straight lines 5x - 2y - 9 = 0 and ay + 2x - 11 = 0…
  8. Find the values of p for which the straight lines 8px + (2 - 3p)y + 1 = 0 and px…
  9. If the straight line passing through the points (h, 3)and (4, 1) intersects the…
  10. Find the equation of the straight line parallel to the line 3x - y + 7 = 0 and…
  11. Find the equation of the straight line perpendicular to the straight line x - 2y…
  12. Find the equation of the perpendicular bisector of the straight line segment…
  13. Find the equation of the straight line passing through the point of…
  14. Find the equation of the straight line which passes through the point of…
  15. Find the equation of the straight line joining the point of intersection of the…
  16. If the vertices of a Δ ABC are A(2, -4), B(3, 3) and C(-1, 5). Find the…
  17. If the vertices of a Δ ABC are A(-4,4), B(8 ,4) and C(8,10). Find the equation…
  18. Find the coordinates of the foot of the perpendicular from the origin on the…
  19. If x + 2y = 7 and 2x + y = 8 are the equations of the lines of two diameters of…
  20. Find the equation of the straight line segment whose end points are the point…
  21. In an isosceles Δ PQR, PQ = PR. The base QR lies on the x-axis, P lies on the…
Exercise 5.6
  1. The midpoint of the line joining (a,- b) and (3a, 5b) isA. (-a, 2b) B. (2a, 4b)…
  2. The point P which divides the line segment joining the points A(1,- 3)and B(-3,…
  3. If the line segment joining the points A(3, 4) and B (14,- 3)meets the x-axis at…
  4. The centroid of the triangle with vertices at (-2,- 5), (-2,12) and (10, -…
  5. If (1, 2), (4, 6), (x, 6)and (3, 2)are the vertices of a parallelogram taken in…
  6. Area of the triangle formed by the points (0,0), (2, 0)and (0, 2)isA. 1 sq.…
  7. Area of the quadrilateral formed by the points (1,1), (0,1), (0, 0)and (1,…
  8. The angle of inclination of a straight line parallel to x-axis is equal toA. 0°…
  9. Slope of the line joining the points (3,- 2)and (-1, a) is - 3/2 , then the…
  10. Slope of the straight line which is perpendicular to the straight line joining…
  11. The point of intersection of the straight lines 9x - y - 2 = 0 and 2x + y - 9 =…
  12. The straight line 4x + 3y - 12 = 0 intersects the y- axis atA. (3, 0) B. (0, 4)…
  13. The slope of the straight line 7y - 2x = 11 is equal toA. - 7/2 B. 7/2 C. 2/7…
  14. The equation of a straight line passing through the point (2 , -7) and parallel…
  15. The x and y-intercepts of the line 2x - 3y + 6 = 0, respectively areA. 2, 3 B.…
  16. The centre of a circle is (-6, 4). If one end of the diameter of the circle is…
  17. The equation of the straight line passing through the origin and perpendicular…
  18. The equation of a straight line parallel to y-axis and passing through the…
  19. If the points (2, 5), (4, 6) and (a, a) are collinear, then the value of a is…
  20. If a straight line y = 2x + k passes through the point (1, 2), then the value…
  21. The equation of a straight line having slope 3 and y-intercept -4 isA. 3x - y -…
  22. The point of intersection of the straight lines y = 0 and x = - 4 isA. (0,- 4)…
  23. The value of k if the straight lines 3x + 6y + 7 = 0 and 2x + ky = 5 are…

Exercise 5.1
Question 1.

Find the midpoint of the line segment joining the points

(i) (1, – 1) and (–5, 3) (ii) (0,0) and (0,4)


Answer:

(i). Midpoint of line–segment joining the points (x1, y1) and (x2, y2)


M (x, y) = M


Midpoint of line–segment joining the points (1, –1) and (–5, 3)


M (x, y) = 



= (–2, 1)


(ii). Midpoint of line–segment joining the points (x1, y1) and (x2, y2)


M (x, y) = M


Midpoint of line–segment joining the points (0, 0) and (0, 4)


M (x, y) = 



= (0, 2)



Question 2.

Find the centroid of the triangle whose vertices are

(i) (1,3), (2, 7) and (12, – 16)

(ii) (3, – 5), (–7, 4) and (10, – 2)


Answer:

i). The centroid G (x, y) of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is given by:


G (x, y) = 


We have (x1, y1) = (1,3), (x2, y2) = (2, 7) and (x3, y3) = (12, – 16)


G (x, y) = 



= (5, –2)


ii). The centroid G (x, y) of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is given by:


G (x, y) = 


We have (x1, y1) = (3, –5), (x2, y2) = (–7, 4) and (x3, y3) = (10, –2)


G (x, y) = 



= (2, –1)



Question 3.

The center of a circle is at (–6, 4). If one end of a diameter of the circle is at the origin, then find the other end.


Answer:

Here, one end of the diameter is at origin that is A (0,0) and let B (a, b) is the required endpoint of the diameter.


Center {O (–6, 4)} of the circle will be exactly middle of the diameter. So, to find another we have to use the mid–point formula


Midpoint of line–segment joining the points (x1, y1) and (x2, y2)


M (x, y) = M


⇒ (–6, 4) = 


⇒ (–6, 4) = 


and 


a = –6 × 2 and b = 4 × 2


a = –12 and b = 8


Therefore, another end point of the diameter (–12, 8)



Question 4.

If the centroid of a triangle is at (1, 3) and two of its vertices are (–7, 6) and (8, 5) then find the third vertex of the triangle.


Answer:

Let A (–7, 8), B (8, 5) and C (a, b) are three vertices of the triangle and centroid of the triangle is (1, 3)


The centroid G (x, y) of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is given by:


G (x, y) = 


⇒ (1, 3) = 


⇒ (1, 3) =


and 


1 + a = 3× 1 and 11 + b = 3 × 3


1 + a = 3 and 11 + b = 9


a = 3 – 1 and b = 9 – 11


a = 2 and b = –2


Therefore, the missing vertex of the triangle is (2, –2).



Question 5.

Using the section formula, show that the points A (1,0), B (5,3), C (2,7) and D (–2, 4) are the vertices of a parallelogram taken in order.


Answer:

The mid–point of diagonals AC and diagonal BD coincide.


Thus, Section Formula internally = 


Where l = 1 and m = 1


Mid–point of diagonal AC


A (1, 0) and C (2, 7)


The mid–point of diagonal is in the ratio of 1:1





Mid–point of diagonal AC


B (5, 3) and D (–2, 4)


The mid–point of diagonal is in the ratio of 1:1





Two diagonals are meeting at the same point. So, the given vertex forms a parallelogram.



Question 6.

Find the coordinates of the point which divides the line segment joining (3, 4) and (–6, 2) in the ratio 3: 2 externally.


Answer:

Section formula externally = 


Where l = 3 and m = 2


A (3, 4) and B (–6, 2)




= (–24, –2)


Therefore, the coordinates of point which divides the line is (–24, –2)



Question 7.

Find the coordinates of the point which divides the line segment joining (–3, 5) and (4, –9) in the ratio 1: 6 internally.


Answer:

Section Formula internally = 


Where l = 1 and m = 6


A (–3, 5) and B (4, –9)





= (–2, 3)


Therefore, the coordinates of point which divides the line is (–2, 3)



Question 8.

Let A (–6, –5) and B (–6, 4) be two points such that a point P on the line AB satisfies AP = AB. Find the point P.


Answer:



⇒ 9 AP = 2 AB


⇒ 9 AP = 2(AP + PB)


⇒ 9AP = 2AP + 2PB


⇒ 9AP – 2AP = 2PB


⇒ 7AP = 2 PB


⇒ 


AP: PB = 7:2


So, P divides the line segment in the ratio is 2:7


Section Formula internally = 


Where l = 2 and m = 7


A (–6, –5) and B (–6, 4)





= (–6, –3)


Therefore, the point P is (–6, –3).



Question 9.

Find the points of trisection of the line segment joining the points A (2, –2) and B (–7, 4).


Answer:

Let P and Q are the points of the intersection of the line segment joining the points A and B.


Here, AP = PQ = QB



AP = 1 PQ = 1 QB = 1


Section Formula internally = 


P divides line segment AB in the ratio 1:2


Where l = 1 and m = 2


A (2, –2) and B (–7, 4)





= (–1,0)


Q divides line segment AB in the ratio 2: 1


Where l = 2 and m = 1


A (2, –2) and B (–7, 4)





= (–4,2)


Therefore, the coordinates of point P (–1, 0) and Q (–4, 2)



Question 10.

Find the points which divide the line segment joining A (–4 ,0) and B (0,6) into four equal parts.


Answer:


Let P, Q and R are the points of the line segment joining the line segment A and B.


Here AP = PQ = QR= RB


AP = 1 PQ = 1 QR = 1 and PB = 1


Section Formula internally = 


P divides line segment AB in the ratio 1:3


Where l = 1 and m = 3


A (–4, 0) and B (0, 6)






Section Formula internally = 


Q divides line segment AB in the ratio 2: 2


Where l = 2 and m = 2


A (–4, 0) and B (0, 6)





= (–2, 3)


Section Formula internally = 


R divides line segment AB in the ratio 3:1


Where l = 3 and m = 1


A (–4, 0) and B (0, 6)






Therefore, the coordinates of point P, Q (–2, 3) and R.



Question 11.

Find the ratio in which the x–axis divides the line segment joining the points (6, 4) and (1, –7).


Answer:

Let l:m be the ratio of the line segment joining the points (6,4) and (1, –7) and let p (x, 0) be the point on x–axis.



Section formula internally: 


(x, 0) = 


(x, 0) = 


Equating the y– coordinates



–7l +4 m = 0


–7l = –4m



l:m = 4: 7


Therefore, x–axis divides the line segment in the ratio 4: 7 internally.



Question 12.

In what ratio is the line joining the points (–5, 1) and (2, 3) divided by the y–axis? Also, find the point of intersection.


Answer:


Let l:m be the ratio of the line segment joining the points (–5,1) and (2, 3) and let C (x, 0) be the point on x–axis.


Section formula internally: 


(x, 0) = 


(x, 0) = 


Equating the x– coordinates



2l – 5m = 0


2l = 5m



l:m = 5: 2


Point of intersection = 




Therefore, y–axis divides the line segment in the ratio 5: 2 internally and point of intersection is .



Question 13.

Find the length of the medians of the triangle whose vertices are (1, –1), (0, 4) and (–5,3).


Answer:

Let A (1, –1), B (0, 4) and C (–5, 3) are the points vertices of triangle.


Let D, E and F are the mid–points of the sides AB, BC and AC respectively.



Mid – point formula = 


Mid – point of AB = 


D = 


D = 


Mid – point of BC = 


E = 


E = 


Mid – point of AC = 


F = 


F = 


F = (–2, 1)


Distance formula = √ (x1– x2)2 + (y1 – y2)2


A (1, –1) and E 


Length of AE =


=






B (0, 4) and F (–2, 1)


Length of BF =


=




A (–5,3) and D


Length of AE =


=








Exercise 5.2
Question 1.

Find the area of the triangle formed by the points

(0, 0), (3, 0) and (0, 2)


Answer:

(0, 0), (3, 0) and (0, 2)


Area of triangle = 


x1 = 0, x2 = 3 and x3 = 0


y1 = 0, y2 = 0 and y3 = 2






= 3sq. units



Question 2.

Find the area of the triangle formed by the points

(5, 2), (3, –5) and (–5, –1)


Answer:

(5, 2), (3, –5) and (–5, –1)


Let A (–5, –1) B (3, –5) and C (5, 2) be the vertices of triangle.


Area of triangle = 


x1 = –5, x2 = 3 and x3 = 5


y1 = –1, y2 = –5 and y3 = 2






= 32 sq. units



Question 3.

Find the area of the triangle formed by the points

(–4, –5), (4, 5) and (–1, –6)


Answer:

(–4, –5), (4, 5) and (–1, –6)


Let A (–4, –5) B (–1, –6) and C (4,5) be the vertices of triangle.


Area of triangle = 


x1 = –4, x2 = –1 and x3 = 4


y1 = –5, y2 = –6 and y3 = 5







= 19 sq. units



Question 4.

Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a.

Vertices: (0, 0), (4, a), (6, 4)

Area (in sq. units): 17


Answer:

Vertices of triangle A (0, 0), B (4, a) and C (6, 4)


Area of triangle = 17 sq. units


Area of triangle = 


x1 = 0, x2 = 4 and x3 = 6


y1 = 0, y2 = a and y3 = 4


⇒ 



⇒ 


⇒ 17 × 2 = 16 – 6a


⇒ 34 = 16 – 6a


⇒ 34 + 6a = 16


⇒ 6a = 16 – 34


⇒ 6a = – 18


⇒ 


⇒ a = –3


Therefore, the required vertices are (4, –3)



Question 5.

Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a.

Vertices: (a, a), (4, 5), (6, –1)

Area (in sq. units): 9


Answer:

Vertices of triangle A (a, a), B (4, 5) and C (6, –1)


Area of triangle = 9 sq. units


Area of triangle = 


x1 = a, x2 = 4 and x3 = 6


y1 = a, y2 = 5 and y3 = –1


⇒ 



⇒ 


⇒ 


⇒ 9 × 2 = 8a – 34


⇒ 18 = 8a – 34


⇒ 8a = 18 +34


⇒ 8a = 52


⇒ 


⇒ 


Therefore, the required vertices are 



Question 6.

Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a.

Vertices: (a, –3), (3, a), (–1,5)

Area (in sq. units): 12


Answer:

Vertices of triangle A (a, –3), B (3, a) and C (–1, 5)


Area of triangle = 12 sq. units


Area of triangle = 


x1 = a, x2 = 3 and x3 = –1


y1 = –3, y2 = a and y3 = 5


⇒ 



⇒ 


⇒ 12 × 2 = a2 – 4a + 27


⇒24 = a2 – 4a + 27


⇒ a2 – 4a + 27 – 24 = 0


⇒ a2 – 4a + 3


⇒ a2 – 3a – a + 3 = 0


⇒ a (a – 3) – (a – 3) = 0


⇒ (a – 3) (a – 1) = 0


a – 3 = 0 or a – 1 = 0


a = 3 or a = 1


Therefore, the required vertices are (3, –3) or (1, –3)



Question 7.

Determine if the following set of points are collinear or not.

(4, 3), (1, 2) and (–2, 1)


Answer:

(4, 3), (1, 2) and (–2, 1)


Let A (4, 3) B (1, 2) and C (–2, 1) be the vertices of triangle.


Area of triangle = 


x1 = 4, x2 = 1 and x3 = –2


y1 = 3, y2 = 2 and y3 = 1


⇒ 


⇒ 


⇒ 


⇒ 


Therefore, the given points are collinear.



Question 8.

Determine if the following set of points are collinear or not.

(–2, –2), (–6, –2) and (–2, 2)


Answer:

(–2, –2), (–6, –2) and (–2, 2)


Let A (–2, –2) B (–6, –2) and C (–2, 2) be the vertices of triangle.


Area of triangle = 


x1 = –2, x2 = –6 and x3 = –2


y1 = –2, y2 = –2 and y3 = 2



⇒ 


⇒ 


⇒ 


Therefore, the given points are non – collinear.



Question 9.

Determine if the following set of points are collinear or not.

, (6, –2) and (–3, 4)


Answer:


Let A  B (6, –2) and C (3, –4) be the vertices of triangle.


Area of triangle = 


, x2 = 6 and x3 = 3


y1 = 3, y2 = –2 and y3 = 4


⇒ 


⇒ 


⇒ 


⇒ 


Therefore, the given points are collinear



Question 10.

In each of the following, find the value of k for which the given points are collinear.

(k, –1), (2, 1) and (4, 5)


Answer:

(k, –1), (2, 1) and (4, 5)


Area of triangle = 


x1 = k, x2 = 2 and x3 = 4


y1 = –1, y2 = 1 and y3 = 5


if points are collinear then, area of triangles are collinear.


⇒ 


⇒ 


⇒ 


⇒ k + 6 – 2 – 5k = 0


⇒4 – 4 k = 0


⇒ 4 = 4k


⇒ 


⇒ k = 1



Question 11.

In each of the following, find the value of k for which the given points are collinear.

(2, – 5), (3, – 4) and (9, k)


Answer:

(2, –5), (3, 4) and (9, k)


Area of triangle = 


x1 = 2, x2 = 3 and x3 = 9


y1 = –5, y2 = –4 and y3 = k


if points are collinear then, area of triangles are collinear.


⇒ 


⇒ 


⇒ 


⇒ 


⇒ k – 2 = 0


⇒ k = 2



Question 12.

In each of the following, find the value of k for which the given points are collinear.

(k, k), (2, 3) and (4, – 1)


Answer:

(k, k) (2, 3) and (4, –1)


Area of triangle = 


x1 = k, x2 = 2 and x3 = 4


y1 = k, y2 = 3 and y3 = –1


⇒ 


⇒ 


⇒ 


⇒ 7k – 2 – k – 12 = 0


⇒ 6k – 14 = 0


⇒ 6k = 14


⇒ 



Question 13.

Find the area of the quadrilateral whose vertices are

(6, 9), (7, 4), (4,2) and (3,7)


Answer:

When the vertices of a quadrilateral is given then its area is given by {(x1–x3)(y2–y4) – (x2–x4)(y1–y3)}


We must take all the vertices in counter clock wise direction otherwise it will give solution in negative.


So, from the figure we assume that


A (x1, y1) = (7,4)


B (x2, y2) = (6, 9)


C (x3, y3) = (3, 7)


D (x4, y4) = (4, 2)



The area of the quadrilateral is {(x1–x3)(y2–y4) – (x2–x4)(y1–y3)}


{(7–3)(9–2) – (6–4)(4–7)}


{28–(–6)} = 17 Sq. units



Question 14.

Find the area of the quadrilateral whose vertices are

(–3, 4), (–5, – 6), (4, – 1) and (1, 2)


Answer:

When the vertices of a quadrilateral is given then its area is given by {(x1–x3)(y2–y4) – (x2–x4)(y1–y3)}


We must take all the vertices in counter clock wise direction otherwise it will give solution in negative.


So, from the figure we assume that


A (x1, y1) = (4, –1)


B (x2, y2) = (1, 2)


C (x3, y3) = (–3, 4)


D (x4, y4) = (–5, –6)



The area of the quadrilateral is {(x1–x3)(y2–y4) – (x2–x4)(y1–y3)}


{(4–(–3))(2–(–6)) –(–1–4)(1–(–5))}


{56+30} = 43 Sq. units



Question 15.

Find the area of the quadrilateral whose vertices are

(–4, 5), (0, 7), (5, – 5) and (–4, – 2)


Answer:

We must take all the vertices in counter clock wise direction otherwise it will give solution in negative.


So, from the figure we assume that


A (x1, y1) = (5, –5)


B (x2, y2) = (0, 7)


C (x3, y3) = (–4, 5)


D (x4, y4) = (–4, –2)



The area of the quadrilateral is {(x1–x3)(y2–y4) – (x2–x4)(y1–y3)}


{(5–(–4))(7–(–2)) –(–5–5)(0–(–4))}


{81+40} = 60.5 Sq. units



Question 16.

If the three points (h, 0), (a, b) and (0, k) lie on a straight line, then using the area of the triangle formula, show that  where h, k ≠ 0.


Answer:

Let A(h,0) B(am) and C (0, k) are the three points


Since the three points A(h,0) B (a, b) and C (0, k) lie on a straight line we can say that the three points are collinear.


So, area of triangle ABC = 0


Area of triangle = 


x1 = h, x2 = a and x3 = 0


y1 = 0, y2 = b and y3 = k


⇒ 


⇒ 


⇒ 


⇒ 0 = hb + ak – kh


⇒ hab + ak = kh


Divided by (kh) on both sides we get


⇒ 


⇒ 


Hence proved.



Question 17.

Find the area of the triangle formed by joining the midpoints of the sides of a triangle whose vertices are (0, – 1), (2,1) and (0,3). Find the ratio of this area to the area of the given triangle.


Answer:

Let A (0, –1), B (2, 1) and C (0, 3) are the vertices of the triangle.


D, E and F are the mid–points of the sides AB, BC and AC respectively.



Mid – point formula = 


Mid – point of AB




D = (1,0)


Mid – point of BC = 





Mid – point of AC = 





Area of Δ ABC:


Area of triangle = 


x1 = 0, x2 = 2 and x3 = 0


y1 = –1, y2 = 1 and y3 = 3


⇒ 


⇒ 


⇒ 


⇒ 


⇒ 4 sq. units


Area of Δ DEF:


Area of triangle = 


x1 = 1, x2 = 1 and x3 = 0


y1 = 0, y2 = 2 and y3 = 1


⇒ 


⇒ 


⇒ 


⇒ 


⇒ 1 sq. units


Area of triangle ABC: Area of triangle DEF


4: 1




Exercise 5.3
Question 1.

Find the angle of inclination of the straight line whose slope is

(i) 1 (ii) √3 (iii) 0


Answer:

If θ is the angle of inclination of the line, then the slope of the line is


m = tan θ where 0° ≤ θ ≤ 180°, θ ≠ 90°


i) tan θ = 1


⇒ θ = 45°


ii) tan θ = √3


⇒ θ = 60°


iii) tan θ = 0


⇒ θ = 0°



Question 2.

Find the slope of the straight line whose angle of inclination is

(i) 30° (ii) 60° (iii) 90°


Answer:

If θ is the angle of inclination of the line, then the slope of the line is m = tan θ


i) Given that m = tan 30°


⇒ 


ii) Given that m = tan 60°


⇒ m = √3


iii) Given that m = tan 30°


⇒ 


∴ The slope is undefined.



Question 3.

Find the slope of the straight line passing through the points

(i) (3, –2) and (7, 2) (ii) (2, –4) and origin

(iii)  and 


Answer:

Slope of straight line passing through the points (x1, y1) and (x2, y2) is given by



i) Slope of straight line passing through the points (3, –2) and (7, 2) is



ii) Slope of straight line passing through the points (2, –4) and (0, 0) is



iii) Slope of straight line passing through the points (1 + √3, 2) and (3 + √3, 4) is




Question 4.

Find the angle of inclination of the line passing through the points

(i) (1, 2) and (2, 3) (ii)  and (0, 0)

(iii) (a, b) and (–a, –b)


Answer:

Slope of straight line passing through the points (x1, y1) and (x2, y2) is given by



i) Slope of straight line passing through the points (1, 2) and (2, 3) is



m = 1


If θ is the angle of inclination of the line, then the slope of the line is


m = tan θ where 0° ≤ θ ≤ 180°, θ ≠ 90°


∴ tan θ = 1 ⇒ θ = 45°


ii) Slope of straight line passing through the points (3, √3) and (0, 0) is



m = √3


If θ is the angle of inclination of the line, then the slope of the line is


m = tan θ where 0° ≤ θ ≤ 180°, θ ≠ 90°


∴ tan θ = √3 ⇒ θ = 30°


iii) Slope of straight line passing through the points (a, b) and (–a, –b) is




If θ is the angle of inclination of the line, then the slope of the line is


m = tan θ where 0° ≤ θ ≤ 180°, θ ≠ 90°


∴ 



Question 5.

Find the slope of the line which passes through the origin and the midpoint of the line segment joining the points (0, – 4) and (8, 0).


Answer:

We need to find slope of a line passing through origin (0, 0) & mid–point of P (0, –4) and Q (8, 0)


Let M mid–point of P (0, –4) and Q (8, 0)


Mid–point formula M (x, y) 


Mid–point of PQ 




We know that,


Slope of line passing through (x1, y1) and (x2, y2) is



Slope of line between points O (0, 0) and M (4, –2)




Hence, slope of line is 



Question 6.

The side AB of a square ABCD is parallel to x–axis. Find the

(i) slope of AB (ii) slope of BC (iii) slope of the diagonal AC


Answer:

i). Slope of AB


Since the side AB is parallel to x–axis,


Slope of side AB = 0


Therefore, the slope of AB is 0


ii). Slope of BC


The angle formed by the side BC is 90°


m = tan θ


θ = 90°


m = tan 90°



Therefore, the slope of BC is not defined.


iii). Slope of the diagonal AC


The diagonal AC is the angle bisector of the angle ∠BAD


∴ θ = 45°


m = tan θ


= tan 45°


= 1


Therefore, the slope of the diagonal AC is 1



Question 7.

The side BC of an equilateral 3ABC is parallel to x–axis. Find the slope of AB and the slope of BC.


Answer:

Slope of line BC:


The side BC is parallel to x–axis.


Slope of the side BC = 0.


Slope of line AB:


Since, this is an equilateral triangle each angle is 60°


∴ θ = 60°


m = tan θ


= tan 60°


= √3


Hence, the slope of AB is √3 and the slope of BC is 0



Question 8.

Using the concept of slope, show that each of the following set of points are collinear.

(2, 3), (3, –1) and (4, –5)


Answer:

(2, 3), (3, –1) and (4, –5)


Let A (2, 3), B (3, –1) and C (4, –5) be the given points


If the given point is collinear then,


Slope of AB = slope of BC


Slope of line passing through (x1, y1) and (x2, y2) is



Slope of AB:




= –4


Slope of BC:




= –4


Slope of AB = Slope of BC = –4


Therefore, the given points are collinear.



Question 9.

Using the concept of slope, show that each of the following set of points are collinear.

(4, 1), (–2, –3) and (–5, –5)


Answer:

(4, 1), (–2, –3) and (–5, –5)


Let A (4, 1), B (–2, –3) and C (–5, –5) be the given points


If the given point is collinear then,


Slope of AB = slope of BC


Slope of line passing through (x1, y1) and (x2, y2) is



Slope of AB:





Slope of BC:






Therefore, the given points are collinear.



Question 10.

Using the concept of slope, show that each of the following set of points are collinear.

(4, 4), (–2, 6) and (1, 5)


Answer:

(4, 4), (–2, 6) and (1, 5)


Let A (4, 4), B (–2, 6) and C (1, 5) be the given points


If the given point is collinear then,


Slope of AB = slope of BC


Slope of line passing through (x1, y1) and (x2, y2) is



Slope of AB:





Slope of BC:






Therefore, the given points are collinear.



Question 11.

If the points (a, 1), (1, 2) and (0, b + 1) are collinear, then show that .


Answer:

Let A (a, 1), B (1, 2) and C (0, b + 1) be the given points.

Slope of line passing through (x1, y1) and (x2, y2) is



Slope of AB 



Slope of BC 



If three points are collinear, then slope of AB is equal to slope of AC


∴ slope of AB = slope of BC



⇒ 1 = (–b + 1)(1 – a)


⇒ 1 = –b(1 – a) + 1(1 – a)


⇒ 1 = –b + ab + 1 – a


⇒ –b + ab – a = 0


⇒ ab – b = a


Dividing both sides by ab, we get


⇒ 


⇒ 


⇒ 



Hence proved.



Question 12.

The line joining the points A (–2, 3) and B (a, 5) is parallel to the line joining the points C (0, 5) and D (–2, 1). Find the value of a.


Answer:

Line joining points A (–2, 3) and B (a, 5) IA parallel to the line joining the points C (0, 5) and D (–2, 1)


The two lines are parallel only when their slopes are equal.


∴ slope of AB = slope of CD


Slope of line passing through (x1, y1) and (x2, y2) is



Slope of AB is:


A (–2, 3) and B (a, 5)




Slope of CD is:


C (0, 5) and D (–2, 1)





As per the property, the two lines are parallel only when their slopes are equal.


i.e. Slope of AB = Slope of CD


⇒ 


⇒ 2 = 2(a + 2)


⇒ 2 = 2a + 4


⇒ 2a = 2 – 4


⇒ 2a = –2


⇒ 


⇒ a = –1



Question 13.

The line joining the points A (0, 5) and B (4, 2) is perpendicular to the line joining the points C (–1, –2) and D (5, b). Find the value of b.


Answer:

Line joining points A (0, 5) and B (4, 2) IA parallel to the line joining the points C (–1, –2) and D (5, b)


The two lines are perpendicular only if the multiplication of their slope is equal to 1.


∴ (Slope of AB) × (Slope of CD) = 1


Slope of line passing through (x1, y1) and (x2, y2) is



Slope of AB is:


A (0, 5) and B (4, 2)




Slope of CD is:


C (–1, –2) and D (5, b)




The two lines are perpendicular only if the multiplication of their slope is equal to 1.


∴ (Slope of AB) × (Slope of CD) = –1


⇒ 


⇒ 


⇒ 


⇒ –b – 2 = –1 × 8


⇒ –b – 2 = –8


⇒ –b = –8 + 2


⇒ – b = – 6


⇒ b = 6



Question 14.

The vertices of ΔABC are A (1, 8), B (–2, 4), C (8, –5). If M and N are the midpoints of AB and AC respectively, find the slope of MN and hence verify that MN is parallel to BC.


Answer:

Given: vertices of triangle ABC i.e. A (1, 8), B (–2, 4), C (8, –5)


M and N are mid – points of AB and AC.


Finding co–ordinates of M and N:


We know that,


M is the mid–point of AB


x1 = 1, x2 = –2


y1 = 8, y2 = 4


Mid–point formula M (x, y) 


Mid–point of AB 




N is the mid–point of AC


x1 = 1, x2 = 8


y1 = 8, y2 = –5


Mid–point of AC 



Slope of MN:



Slope of line passing through (x1, y1) and (x2, y2) is








Verification of MN and BC are parallel:


If MN and BC are parallel, then their slopes must be equal.


Slope of BC:


B (–2, 4) and C (8, –5)


Slope of BC 




∴ Slope of MN = Slope of BC = 


Hence, MN is parallel to BC.



Question 15.

A triangle has vertices at (6, 7), (2, –9) and (–4, 1). Find the slopes of its medians.


Answer:

Given: Vertices of triangle A (6, 7), B (2, –9) and C (–4, 1)


To find the slopes of medians. We need to know the mid–points of AB, BC and AC.


D, E and F are mid–points of AB, BC and AC respectively.



Mid – point formula = 


Mid – point of AB = 





Mid – point of BC = 





Mid – point of AC = 




F = (1, 4)


Slopes of median of triangles:


Slope of line passing through (x1, y1) and (x2, y2) is



A (6, 7) and E (–1, –4)


Slope of AE




B (2, –9) and E (1, 4)


Slope of BF




C (–4, 1) and D (4, –1)


Slope of CD





Question 16.

The vertices of a ΔABC are A (–5, 7), B (–4, –5) and C (4, 5). Find the slopes of the altitudes of the triangle.


Answer:

Let AD, BE and CF be the altitudes of a ΔABC.



Since, the altitude AD is perpendicular to BC,


Slope of BC 




(slope of BC) × (Slope of AD) = –1 (∵ m1m2 = –1)


Let slope of AD be m1.


⇒ 


⇒ 


Since, the altitude BE is perpendicular to AC,


Slope of AC 




(slope of AC) × (Slope of BE) = –1 (∵ m1m2 = –1)


Let slope of BE be m2.


⇒ 


⇒ 


Since, the altitude CF is perpendicular to AB,


Slope of AB 




(slope of AB) × (Slope of CF) = –1 (∵ m1m2 = –1)


Let slope of CF be m3.


⇒ 


⇒ 



Question 17.

Using the concept of slope, show that the vertices (1, 2), (–2, 2), (–4, –3) and (–1, –3) taken in order form a parallelogram.


Answer:

Let A (–2, 2), B (1, 2), C (–1, –3) and D (–4, –3) be the given points taken in order.



Now,


Slope of line passing through (x1, y1) and (x2, y2) is



Slope of AB 



= 0


Slope of CD = 




= 0


∴ Slope of AB = Slope of CD


Hence, AB is parallel to CD. … (1)


Now,


Slope of BC 




Slope of AD = 





∴ Slope of BC = Slope of AD


Hence, BC is parallel to AD. … (2)


From (1) and (2), we see that opposite sides of quadrilateral are parallel.


∴ ABCD is a parallelogram.



Question 18.

Show that the opposite sides of a quadrilateral with vertices A (–2, –4), B (5, –1), C (6, 4) and D (–1, 1) taken in order are parallel.


Answer:

Let A (–2, 2), B (1, 2), C (–1, –3) and D (–4, –3) be the given points taken in order.



Now,


Slope of line passing through (x1, y1) and (x2, y2) is



Slope of AB 




Slope of CD = 




∴ Slope of AB = Slope of CD


Hence, AB is parallel to CD. … (1)


Now,


Slope of BC 




Slope of AD = 





∴ Slope of BC = Slope of AD


Hence, BC is parallel to AD. … (2)


From (1) and (2), we see that opposite sides of quadrilateral are parallel.




Exercise 5.4
Question 1.

Write the equations of the straight lines parallel to x- axis which are at a distance of 5 units from the x–axis.


Answer:

Given


The line is ∥ to x axis that means it is a horizontal line at the fixed distance from x axis.


The equation of this line would be of the form y = k where k is some constant.


Here it is given the distance from x axis is 5 units which could be either in positive or negative direction, so the required equation would be


y = 5 or y = –5



Question 2.

Find the equations of the straight lines parallel to the coordinate axes and passing through the point (–5,–2).


Answer:

A line parallel to x axis is horizontal line and when this line passes through the given point (–5,–2) the equation of the line would be y = –2


When the line is parallel to the y axis, it is a vertical straight line passing through the point (–5, –2), the required equation of the line would be x = –5



Question 3.

Find the equation of a straight line whose

(i) slope is –3 and y–intercept is 4. (ii) angle of inclination is 60° and y–intercept is 3.


Answer:

The equation of the straight line with the given slope (m) and y –intercept ‘c’ is given the slope–intercept form


i.e. y = mx + c


(i) Here given slope = –3 (m) and y intercept = 4 (c)


So the required equation is


y = –3x + 4


⇒ 3x + y–4 = 0


(ii) Given angle of inclination = 60° and y intercept = 3 (c)


Here slope of the line (m) = tan θ = tan 60° = √3


The required equation is


y = mx + c


⇒ y = √3x + 3


⇒ √3x –y + 3 = 0



Question 4.

Find the equation of the line intersecting the y- axis at a distance of 3 units above the origin and tan , where θ is the angle of inclination.


Answer:

Given


Angle of inclination = tan θ = 1/2 = slope of the line (m)


Also, the y intercept = 3 units (c) as the line is intersecting y axis at a distance of 3 units above origin


The equation of the straight line with the given slope (m) and y –intercept ‘c’ is given the slope–intercept form


i.e. y = mx + c



⇒ 2y = x + 6


⇒ x – 2y + 6 = 0


x – 2y + 6 = 0



Question 5.

Find the slope and y–intercept of the line whose equation is

(i) y = x + 1 (ii) 5x = 3y (iii) 4x – 2y + 1 = 0 (iv) 10x + 15y + 6 = 0


Answer:

The equation of the straight line with the slope (m) and y –intercept ‘c’ in the slope–intercept form is


y = mx + c


(i) here y = x + 1


⇒ slope of the line (m) = 1 and y–intercept ( c ) = 1


(ii) here 5x = 3y


3y = 5x


⇒ 


Thus the slope of the line =  and y intercept ( c) = 0


(iii) 4x– 2y + 1 = 0


⇒ –2y = –4x –1


⇒ 


Thus the slope of the line (m) = 2


and y –intercept (c) = 1/2


(iv) 10x + 15y + 6 = 0


⇒ 15y = = 10x – 6


⇒ 


Thus the slope of the line (m) =  and y –intercept (c) = 



Question 6.

Find the equation of the straight line whose

(i) slope is –4 and passing through (1, 2) (ii) slope is and passing through (5, –4)


Answer:

The equation of the line with the given slope m and passing through the point (x1, y1) is given by the slope–point form


y– y1 = m (x – x1)


(i) Here given the slope (m) = –4 and the point = (1,2)


Thus the equation of the line is


y–2 = –4(x – 1)


⇒ y –2 = –4x + 4


⇒ y = –4x + 6


⇒ 4x + y – 6 = 0


(ii) Here given slope (m) = 2/3 and point = (5,–4)


Thus the required equation is



⇒ 3y + 12 = 2x –10


⇒ –2x + 3y + 22 = 0


⇒ 2x – 3y – 22 = 0



Question 7.

Find the equation of the straight line which passes through the midpoint of the line segment joining (4, 2) and (3, 1) whose angle of inclination is 30°.


Answer:

The equation of the line with the given slope m and passing through the point (x1, y1) is given by the slope–point form


y– y1 = m (x – x1)


here given angle of inclination = 30°


⇒ slope (m) = tan 30 ° = 


Also the line passes through the midpoint of the line segment joining points (4, 2) and (3, 1)


⇒ the line passes through the point ( 


The equation of the required line is




⇒ 


⇒ 2x –2 √3 y + (3√3–7) = 0



Question 8.

Find the equation of the straight line passing through the points

(i) (–2, 5) and (3, 6) (ii) (0, –6) and (–8, 2)


Answer:

The equation of the line passing through the two given (x1, y1) and (x2,y2)points is given by form



(i) Here, (–2,5) and (3,6) are tow given points


The required equation is



⇒ 


⇒ 5y –25 = x + 2


⇒ x–5y + 27 = 0


(ii) Here two points are (0, –6) and (–8, 2)


The equation is given by



⇒ 


⇒ –8y – 48 = 8x


⇒ x + y + 6 = 0 (dividing the whole equation by 8)



Question 9.

Find the equation of the median from the vertex R in a ΔPQR with vertices at

P(1, –3), Q(–2, 5) and R(–3, 4).


Answer:

The given points are


P (1,–3), Q(–2,5) and R (–3,4)


The median from R will pass through the midpoint of line PQ


(∵ Median passes through the mid–point of the side opposite to the vertex)


Let the mid–point of PQ be A = 


The equation of the line passing through the two given (x1, y1) and (x2,y2)points is given by form



Thus here the required equation is



⇒ 


⇒ 


⇒ 


⇒ 5y –20 = –6x –18


⇒ 6x + 5y –2 = 0



Question 10.

By using the concept of the equation of the straight line, prove that the given three points are collinear.

(i) (4, 2), (7, 5) and (9, 7)

(ii) (1, 4), (3, –2) and (–3, 16)


Answer:

The equation of the line passing through the two given (x1, y1) and (x2,y2)points is given by form



(i) The equation of the line passing through (4, 2), (7, 5)is



⇒ y–2 = x–4


⇒ x – y –2 = 0


Substituting the 3rd point in the equation of the line obtained


x–2 = y


⇒ 7 = 9–2


⇒ 7 = 7


Hence LHS = RHS


Since the third point satisfies the equation of the line obtained


⇒ All the three points lies in the same straight line or are collinear.


(ii) (1, 4), (3, –2) and (–3, 16)


The equation of the line passing through the points (1, 4), (3, –2) is



⇒ 


⇒ y–4 = –3x + 3


⇒ 3x + y –7 = 0


Substituting the 3rd point in the equation of line obtained


y = –3x + 7


⇒ 16 = –3 (–3) + 7


⇒ 16 = 16


LHS = RHS


Thus the three points are in the same straight line or are collinear.



Question 11.

Find the equation of the straight line whose x and y–intercepts on the axes are given by

(i) 2 and 3 (ii) 

(iii) 


Answer:

The equation of the straight line whose x and y intercepts are given as ‘a’ and ‘b’ respectively, is of the form



(i) Here given x intercept as 2 and y intercept as 3


The equation of the line



⇒ 3x + 2y = 6


⇒ 3x + 2y –6 = 0


(ii) Given x intercept as  and y intercept as 


The equation of the line


⇒ 


⇒ 


⇒ –9x + 2y = 3


⇒ –9x + 2y –3 = 0


(iii) Given x intercept as and y intercept as 


The equation of the line


⇒ 


⇒ 


⇒ 


⇒ 15x –8y –6 = 0



Question 12.

Find the x and y intercepts of the straight line

(i) 5x + 3y – 15 = 0 (ii) 2x – y + 16 = 0

(iii) 3x + 10y + 4 = 0


Answer:

The equation of the straight line whose x and y intercepts are given as ‘a’ and ‘b’ respectively, is of the form



(i) The given equation is


5x + 3y – 15 = 0


⇒ 5x + 3y = 15


⇒ 


⇒ 


Hence the x intercept is 3 and y intercept is 5


(ii) The given equation is


2x – y + 16 = 0


⇒ 2x– y = –16


⇒ 


⇒ 


Hence the x intercept is –8 and y intercept is 16.


(iii) Given equation is


3x + 10y + 4 = 0


⇒ 3x + 10y = –4


⇒ 


⇒ 


⇒ 


Hence the x intercept is  and y intercept is 



Question 13.

Find the equation of the straight line passing through the point (3, 4) and has intercepts which are in the ratio 3 : 2.


Answer:

Given that the intercepts are in the ratio 3:2


Let the x–intercept be 3k and 2k, where k is some constant


Using the intercept –form, the equation of the line is



⇒ 


⇒ 


The point (3, 4) lies on this line, thus the point will satisfy the equation of the line


⇒ 


⇒ 


⇒ 1 + 2 = k


⇒ K = 3


Thus the equation of the line is


⇒ 


⇒ 2x + 3y = 18


⇒ 2x + 3y –18 = 0



Question 14.

Find the equation of the straight lines passing through the point (2, 2) and the sum of the intercepts is 9.


Answer:

Let the x intercept be ‘a’ and the y intercept be ‘b’ . it is given that


a + b = 9


The equation of the line using the intercept form is



Substituting b = 9– a in this equation



⇒ 


⇒ x (9–a) + ya = a(9 –a)


The point (2, 2) lies on this line and thus it satisfies the equation


x (9–a) + ya = a(9 –a)


⇒ 2(9 –a) + 2a = a ( 9–a)


⇒ 18 –2a + 2a = 9a –a2


⇒ a2 – 9a + 18 = 0


⇒ a2 – 3a –6a + 18 = 0


⇒ a ( a– 3) –6 (a –3) = 0


⇒ (a–6) (a–3) = 0


⇒ a = 6 or 3


The equation of the line is


x( 9–3) + y(3) = 3(9–3)


⇒ 6x + 3y = 18


⇒ 2x + y –6 = 0


Or


x(9–6) + y(6) = 6(9–6)


3x + 6y = 18


⇒ x + 2y –6 = 0



Question 15.

Find the equation of the straight line passing through the point (5, –3) and whose intercepts on the axes are equal in magnitude but opposite in sign.


Answer:

Let the x intercept be ‘a’. it is given the y intercept is equal in magnitude but opposite in sign


So y intercept = ‘–a’


The equation of the line using the intercept form is



⇒ 


⇒ 


⇒ x–y = a


The point (5, –3) lie son this line, thus it satisfies the given equation of the line


5 –(–3) = a


⇒ a = 8


The equation of the line is


x– y – 8 = 0



Question 16.

Find the equation of the line passing through the point (9, –1) and having its x–intercept thrice as its y–intercept.


Answer:

Let the x intercept be ‘a’. it is given that x intercept is thrice the y intercept


⇒ y intercept = 


The equation of the line using the intercept form is



Substituting b = a/3 in the equation



⇒ 


⇒ x + 3y = a


the point (9,–1) lies on the equation of the line thus it must satisfies it


⇒ 9 + 3(–1) = a


⇒ a = 6


Hence the equation of the line is


x + 3y –6 = 0



Question 17.

A straight line cuts the coordinate axes at A and B. If the midpoint of AB is (3, 2), then

find the equation of AB.


Answer:

The mid–point of AB is in the 1st quadrant as both the intercepts are positive.


Thus A will lie on the y –axis and B will lie on the x axis


Let A be ( 0,a) and B be (b,0)


By the mid–point theorem, the coordinates of mid–point are


( = (3,2)


⇒ 


⇒  and


⇒ B = 6 and a = 4


Thus y intercept is 4 and x intercept is 6


The equation of the line using the intercept form is



⇒ 


⇒ 4x + 6y = 24


⇒ 2x + 3y –12 = 0



Question 18.

Find the equation of the line passing through (22, –6) and having intercept on x–axis exceeds the intercept on y–axis by 5.


Answer:

Let the x intercept be ‘a’. It is given that x intercept exceeds the y intercept by 5


⇒ y intercept = a– 5


The equation of the line using the intercept form is



Substituting Value of b




⇒ x (a–5) + ya = a(a–5)


Given that the point (22, –6) lies on this equation of the line, hence it should satisfy it


22(a–5) + (–6) a = a(a–5)


⇒ 22a – 110 – 6a = a2 –5a


⇒ a2 – 21a + 110 = 0


⇒ a(a–10) – 11(a –10) = 0


⇒ (a–11) (a–10)


⇒ a = 11 or 10


The equation of the line is


x(11–5) + 11y = 11(11–5)


⇒ 6x + 11y – 66 = 0


Or


x(10–5) + 10y = 10(10–5)


⇒ 5x + 10y– 50 = 0


⇒ x + 2y –10 = 0



Question 19.

If A(3, 6) and C(–1, 2) are two vertices of a rhombus ABCD, then find the equation of straight line that lies along the diagonal BD.


Answer:

In a rhombus, the diagonals are perpendicular to each other and also bisect each other


Thus, the product of the slope of AC and BD will be –1 and BD will pass through the mid–point of AC


The points are given as A (3,6) and C ( –1, 2)


Slope of AC is



Thus, the slope of BD (m) = –1


The mid–point of AC is



Thus BD passes through the point (1,4)


Now using the slope point form, the equation of the line BD is:


(y –y1) = m (x – x1)


Here m = 1 and point is (1, 4)


⇒ y – 4 = (–1) (x –1)


⇒ y – 4 = –x + 1


⇒ x + y –5 = 0



Question 20.

Find the equation of the line whose gradient is  and which passes through P, where P divides the line segment joining A(–2, 6) and B (3, –4) in the ratio 2 : 3 internally.


Answer:

According to the section formula, the coordinates of the point P (x, y) with the given ratio m1 : m2 is


P = 





The gradient is the slope, thus the slope of the line is given as 


Using the slope –point form, the equation of the line is


y– y1 = m (x– x1)



⇒ 2y – 4 = 3x


⇒ 3x – 2y + 4 = 0




Exercise 5.5
Question 1.

Find the slope of the straight line

3x + 4y – 6 = 0


Answer:

Here we have the straight line : 3x + 4y –6 = 0


The slope intercept form is y = mx + b, where m is the slope and b is the y intercept.


Therefore, 3x + 4y –6 = 0


⇒ 4y + 3x = 6


⇒ 4y = 6–3x


⇒  (Divide both sides of the equation by 4)


⇒ 


Now we will rewrite it in the slope intercept form



Hence according to the slope intercept form, y = mx + b


m i.e. slope is .



Question 2.

Find the slope of the straight line

y = 7x + 6


Answer:

Here we have the straight line : y = 7x + 6


The slope intercept form is y = mx + b, where m is the slope and b is the y intercept.


Therefore when, y = 7x + 6


Hence according to the slope intercept form,


y = mx + b


m i.e. slope is 7.



Question 3.

Find the slope of the straight line

4x = 5y + 3.


Answer:

Here we have the straight line : 4x = 5y + 3


The slope intercept form is y = mx + b, where m is the slope and b is the y intercept.


Therefore, 4x = 5y + 3


⇒ –5y – 3 = –4x (Commutative law)


⇒ 5y + 3 = 4x (multiply by – on both sides of the equation)


⇒  (Divide both sides of the equation by 5)


⇒ 


⇒ 


Now we will rewrite it in the slope intercept form



Hence according to the slope intercept form, y = mx + b


m i.e. slope is .



Question 4.

Show that the straight lines x + 2y + 1 = 0 and 3x + 6y + 2 = 0 are parallel.


Answer:

Given: Here the straight lines are x + 2y + 1 = 0 and


3x + 6y + 2 = 0.


To Prove: x + 2y + 1 = 0 and 3x + 6y + 2 = 0 are parallel.


Proof: If two lines are parallel then their slopes are equal.


Here slope of the first line x + 2y + 1 = 0 will be



(When the line is in the form ax + by + c = 0 then the slope of the line is )


⇒ 


Here slope of the second line 3x + 6y + 2 = 0 will be



⇒ 


⇒ 


Now both the slopes are equal.


Hence both the lines are parallel.



Question 5.

Show that the straight lines 3x – 5y + 7 = 0 and 15x + 9y + 4 = 0 are perpendicular.


Answer:

Given: The straight lines are 3x – 5y + 7 = 0 and


15x + 9y + 4 = 0 .


To Prove: The straight lines are 3x – 5y + 7 = 0 and


15x + 9y + 4 = 0 are perpendicular.


Proof: If two lines are perpendicular, then the product of their slopes is equal to –1.


The slope of the first line,3x –5y + 7 = 0 is


 ()


The slope of the second line, 15x + 9y + 4 = 0


is  ()


⇒ 


Now the product of these slopes is m1×m2.


⇒ 


As the product of the slopes is –1, the lines are perpendicular to


each other.



Question 6.

If the straight lines  and ax + 5 = 3y are parallel, then find a.


Answer:

Given: The straight lines are  and ax + 5 = 3y are parallel.


Since the lines are parallel their slopes should be equal.


Now,


⇒ y = 2(x–p)


⇒ y = 2x –2p


⇒ m1 = 2


ax + 5 = 3y


⇒ 3y = ax + 5


⇒ 


⇒ 


⇒ 


Now as the lines are parallel ,m1 = m2


⇒ 


⇒ 2×3 = a


⇒ a = 6



Question 7.

Find the value of a if the straight lines 5x – 2y – 9 = 0 and ay + 2x – 11 = 0 are perpendicular to each other.


Answer:

Given: The straight lines 5x – 2y – 9 = 0 and ay + 2x – 11 = 0 are perpendicular to each other.


As the lines are perpendicular to each other, the product of their slopes is equal to –1.


Slope of the first line 5x – 2y – 9 = 0 is m1.


 ()


Slope of the second line ay + 2x –11 = 0 is m2.



Therefore,


⇒ 


As the lines are perpendicular m1× m2 = –1


⇒ 


⇒ a = 5



Question 8.

Find the values of p for which the straight lines 8px + (2 – 3p)y + 1 = 0 and px + 8y – 7 = 0 are perpendicular to each other.


Answer:

Given: The straight lines 8px + (2 – 3p)y + 1 = 0 and


px + 8y – 7 = 0 are perpendicular to each other.


Since the lines are perpendicular to each other, product of their slopes is equal to –1.


Slope of the first line 8px + (2 – 3p)y + 1 = 0 is m1.


i.e. 


Slope of the second line px + 8y – 7 = 0 is m2.


i.e. 


As the lines are perpendicular to each other m1× m2 = –1.


⇒ 


⇒ 


⇒ p2 = –1×(2–3p)


⇒ p2 = 3p–2


⇒ p2–3p + 2 = 0


⇒ p.p–2p–p + 2 = 0


⇒ p(p–2)–1(p–2) = 0


⇒ (p–2)(p–1) = 0


⇒ p–2 = 0 ,p–1 = 0


⇒ p = 2,p = 1


Hence p = 1,2.



Question 9.

If the straight line passing through the points (h, 3)and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle, then find the value of h.


Answer:

Given: The straight line passing through the points (h, 3)


and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle.


As these lines intersect at right angle, they are perpendicular to each other.


When the lines are perpendicular to each other, the product of their slopes is equal to –1.


Slope of the first line passing through the points (h,3) and (4,1)


 (here the two points are(x1,y1) and (x2,y2))


Now, 


⇒ 


Slope of the second line 7x – 9y –19 = 0 is



Therefore, product of the slopes is 


⇒ 


⇒ 


⇒ 


⇒ 14 = 36–9h


⇒ 9h = 36–14 = 22


⇒ 



Question 10.

Find the equation of the straight line parallel to the line 3x – y + 7 = 0 and passing through the point (1, –2).


Answer:

Here it’s given that the straight line is parallel to the line


3x – y + 7 = 0 and passing through the point (1, –2).


As the lines are parallel to each other their slopes are equal.


Slope of the given line 3x – y + 7 = 0 is



Equation of the line passing through the point(1,–2) is


(y–y1) = m(x–x1),where (x1,y1) is (1,–2)


⇒ (y–(–2)) = 3(x–1)


⇒ y + 2 = 3x–3


⇒ y + 2–3x + 3 = 0


⇒ –3x + y + 5 = 0


⇒ 3x–y–5 = 0 (multiplied by –1 on both sides of the equation)


Hence the equation of the line is 3x–y–5 = 0 .



Question 11.

Find the equation of the straight line perpendicular to the straight line x – 2y + 3 = 0 and passing through the point (1, –2).


Answer:

Here it’s given that the straight line is perpendicular to the


straight line x – 2y + 3 = 0 and passing through the point (1, –2).


As the lines are perpendicular to each other , the product of their


slopes is equal to –1.


Slope of the given line x–2y + 3 = 0 is



Equation of the line passing through the point(1,–2) is


(y–y1) = m(x–x1),where (x1,y1) is (1,–2)


)


⇒ 2(y + 2) = (x–1)


⇒ 2y + 4 = x–1



Question 12.

Find the equation of the perpendicular bisector of the straight line segment joining the points (3, 4) and (–1, 2).


Answer:

Given: There is a perpendicular bisector of the straight line segment joining the points (3, 4) and (–1, 2).


We have to find the equation of the perpendicular bisector.


As it is perpendicular to the given line segment,the product of their slopes is equal to –1 and as it bisects the line segment,it implies it divides the line segment into 2 equal parts.


Thus,mid–point of the line segment joining the points (3, 4) and (–1, 2) is:


; where (x1,y1) and (x2,y2) are the end points of the line segment





= 1,3


Therefore the mid–point is (1,3).


The slope of the line segment joining the points (3, 4) and


(–1, 2) is:



⇒ 


⇒ 


Therefore the equation of the perpendicular bisector is


(y–y1) = m(x–x1)


Now substitute the value of the mid–point(1,3) and slope in the above equation.



⇒ 2(y–3) = (x–1)


⇒ 2y –6 = x–1


⇒ 2y–6–x + 1 = 0


⇒ –x + 2y–5 = 0


⇒ x–2y + 5 = 0 (multiply by –1 on both the sides of the equation)



Question 13.

Find the equation of the straight line passing through the point of intersection of the lines 2x + y – 3 = 0 and 5x + y – 6 = 0 and parallel to the line joining the points (1, 2) and (2, 1).


Answer:

Here it is given that the straight line passing through the point of intersection of the lines 2x + y – 3 = 0 and 5x + y – 6 = 0 and parallel to the line joining the points (1, 2) and (2, 1).


As the straight line passes through the point of intersection of the lines 2x + y – 3 = 0 and 5x + y – 6 = 0 , we should find the intersection point by solving these equations:



i.e –3x = –3


⇒ x = 1


Substitute x = 1 in 2x + y–3 = 0


We get,2(1) + y–3 = 0


⇒ y–1 = 0


⇒ y = 1


Therefore , the intersection point is (1,1).


As the line passing through(1,1) is parallel to the line segments


joining the points (1, 2) and(2, 1),their slopes are equal.


Slope of the line joining the points (1, 2) and(2, 1) is



⇒ 


Hence the equation of the line passing through the point (1,1) with slope m equal to –1 is


(y–y1) = m(x–x1)


⇒ (y–1) = –1(x–1)


⇒ y–1 = –x + 1


⇒ y–1 + x–1 = 0


⇒ x + y–2 = 0



Question 14.

Find the equation of the straight line which passes through the point of intersection of the straight lines 5x – 6y = 1 and 3x + 2y + 5 = 0 and is perpendicular to the straight line 3x – 5y + 11 = 0.


Answer:

Here we have the straight line which passes through the point of intersection of the straight lines 5x – 6y = 1 and 3x + 2y + 5 = 0 and is perpendicular to the straight line 3x – 5y + 11 = 0.


As the straight line passes through the point of intersection of the lines 5x – 6y = 1 and 3x + 2y + 5 = 0, we should find the intersection point by solving these equations:


5x –6y = 1


3x + 2y + 5 = 0


⇒ 5x – 6y– 1 = 0 –––(1)


and 3x + 2y + 5 = 0–––(2)


Now multiply equation (2) by 3 on both the sides.


Thus we have 9x + 6y + 15 = 0–––(3)


Now, we have



⇒ 14x = –14


⇒ x = –1


Now, substituting x = –1 in the equation 5x–6y = 1, we have,


5(–1)–6y = 1


⇒ –5–6y = 1


⇒ –6y = 1 + 5 = 6


⇒ 


⇒ y = –1


Therefore the point of intersection is (–1,–1).


Slope of the line 3x –5y + 11 = 0 is :



As the line which passes through (–1,–1) is perpendicular to 3x –5y + 11 = 0 ,the product of their slopes will be –1.


Therefore,


⇒  (Here we have multiplied  on both the sides)


⇒ 


Hence the equation of the line passing through (–1,–1) and slope as is:


(y–y1) = m(x–x1)


⇒ 


⇒ 


⇒ 3y + 3 = –5x–5


⇒ 3y + 3 + 5x + 5 = 0


⇒ 5x + 3y + 8 = 0



Question 15.

Find the equation of the straight line joining the point of intersection of the lines 3x – y + 9 = 0 and x + 2y = 4 and the point of intersection of the lines 2x + y – 4 = 0 and x – 2y + 3 = 0.


Answer:

Here we have the straight line which joins the point of intersection of the lines 3x – y + 9 = 0 and x + 2y = 4 and the point of intersection of the lines 2x + y – 4 = 0 and x – 2y + 3 = 0.


As the straight line which joins the point of intersection of the lines 3x – y + 9 = 0 and x + 2y = 4,let us solve these 2 equations:


3x –y + 9 = 0–––(1)


X + 2y–4 = 0–––(2)


⇒ 3x –y + 9 = 0–––(1)


3x + 6y–12 = 0–––(2)(multiply (2) equation by 3 on both the sides)


Now,


(Subtract equation (2) from (1))



⇒ –7y = –21


⇒ 


Substitute y = 3 in the first equation


3x–y + 9 = 0


⇒ 3x–3 + 9 = 0


⇒ 3x + 6 = 0


⇒ 3x = –6


⇒ 


Thus, the point of intersection is(–2,3).


Now,


2x + y – 4 = 0–––(3)


x – 2y + 3 = 0–––(4)


(multiply (4) equation by 2 on both the sides) we get


2x – 4y + 6 = 0 --------(5)


Now,



⇒ 5y = 10


⇒ 


Substitute y = 2 in the (3) equation:


2x + y–4 = 0


⇒ 2x + 2–4 = 0


⇒ 2x–2 = 0


⇒ 2x = 2


⇒ 


The point of intersection is (1,2).


Hence equation of the line is



Here we have the points (–2,3) and (1,2)


Therefore,





⇒ 3(y–3) = –1(x + 2)


⇒ 3y–9 = –x–2


⇒ 3y–9 + x + 2 = 0


⇒ 3y + x–7 = 0


⇒ x + 3y–7 = 0



Question 16.

If the vertices of a Δ ABC are A(2, –4), B(3, 3) and C(–1, 5). Find the equation of the straight line along the altitude from the vertex B.


Answer:


Given: A ∆ABC has vertices A(2,–4),B(3,3) and C(–1,5).


The straight line BD drawn from vertex B is perpendicular to the line AC. So the product of the slopes of BD and AC is equal to –1.


Slope of AC is m1.



Where (x1,y1) and (x2,y2) are (2,–4) and (–1,5)


Therefore,


Slope of the BD is 



Hence the equation of the line BD drawn from the vertex B is


(y–y1) = m(x–x1),here B is (3,3) and m = 


⇒ 


⇒ 3(y–3) = (x–3)


⇒ 3y –9 = x–3


⇒ 3y–9–x + 3 = 0


⇒ –x + 3y–6 = 0


⇒ x–3y + 6 = 0



Question 17.

If the vertices of a Δ ABC are A(–4,4 ), B(8 ,4) and C(8,10). Find the equation of the straight line along the median from the vertex A.


Answer:


Given: The ∆ABC with vertices A(–4,4),B(8,4) and C(8,10) and


AD is the median,i.e. it passes through the mid point of BC.


Therefore D is midpoint of BC where B(8,4) and C(8,10).






Equation of AD is


,where (x1,y1) and (x2,y2) are (–4,4) and (8,7)


⇒ 


⇒ 


⇒ 


⇒ 12(y–4) = 3(x + 4)


⇒ 12y–48 = 3x + 12


⇒ 12y–48–3x–12 = 0


⇒ –3x + 12y–60 = 0


⇒ 3x–12y + 60 = 0


⇒ x–4y + 15 = 0(Divide both sides of the equation by 3)


Hence Equation of AD is x–4y + 15 = 0.



Question 18.

Find the coordinates of the foot of the perpendicular from the origin on the straight line 3x + 2y = 13.


Answer:

Here we have a perpendicular from the origin i.e.(0,0) to the


straight line 3x + 2y = 13.


We have to find the foot of the perpendicular i.e the intersection point at the line 3x + 2y = 13.


As these lines are perpendicular the product of their slopes is equal to –1.


Slope of the 3x + 2y–13 = 0 is m.



Therefore the slope of perpendicular is 


i.e.


Hence the equation of the perpendicular from (0,0) and slope as 


is (y–y1) = m(x–x1)



⇒ 


⇒ 3y = 2x


⇒ 3y–2x = 0


⇒ 2x–3y = 0


Now solve the two equations 3x + 2y–13 = 0 and 2x–3y = 0.


3x + 2y–13 = 0–––(1)


2x–3y = 0–––––(2)


Multiply (1) by 3 and (2) by 2 and add



⇒ 


Substitute x = 3 in the equation 2x–3y = 0.


2(3)–3y = 0


⇒ –3y = –6


⇒ y = 2(Divide both the sides of the equation by –3)


Hence the coordinates of the foot of the perpendicular is(3,2)



Question 19.

If x + 2y = 7 and 2x + y = 8 are the equations of the lines of two diameters of a circle, find the radius of the circle if the point (0, –2) lies on the circle.


Answer:

Given: The equations of the lines of two diameters of a circle


are x + 2y = 7 and 2x + y = 8 and F(0,–2)



Now to find the center of the circle,we have to find the


intersection of the lines x + 2y = 7 and 2x + y = 8


x + 2y = 7–––(1)


2x + y = 8–––(2)


Multiplying (1) by 2,we get




Substitute y = 2 in x + 2y = 7 we get,


x + 2(2) = 7


⇒ x = 7 – 4 = 3


The point of intersection is (3,2) i.e.the center.


Therefore distance between the points (3,2) and (0,–2) is






= 5 units


Hence the radius of the circle is 5 units.



Question 20.

Find the equation of the straight line segment whose end points are the point of intersection of the straight lines

2x – 3y + 4 = 0, x – 2y + 3 = 0 and the midpoint of the line joining the points (3, –2) and (–5, 8).


Answer: 


Here its given that the straight line segment has end points as the point of intersection of the straight lines


2x – 3y + 4 = 0, x – 2y + 3 = 0 and the midpoint of the line joining the points (3, –2) and (–5, 8).


As one end point is point of intersection of the straight lines


2x – 3y + 4 = 0, x – 2y + 3 = 0,we will solve these equations.


2x–3y + 4 = 0––––(1)


x–2y + 3 = 0–––––(2)


Multiply (2) by 2,then we have



i.e. y = 2


Substitute y = 2 in 2x–3y + 4 = 0


2x–3(2) + 4 = 0


⇒ 2x–6 + 4 = 0


⇒ 2x–2 = 0


⇒ 2x = 2


⇒ x = 1


Therefore the lines intersect at (1,2).


Now the line has one end point as (1,2) and other end point as mid– point of the line joining (3, –2) and (–5, 8).


The mid– point of (3, –2) and (–5, 8) is





= (–1,3)


Hence to find the equation of the line with end points as (x1,y1) and (x2,y2) ,we use:


,where (x1,y1) and (x2,y2) are (1,2) and (–1,3)


⇒ 


⇒ 


⇒ –2(y–2) = (x–1)


⇒ –2y + 4 = x–1


⇒ –2y + 4–x + 1 = 0


⇒ –2y–x + 5 = 0


⇒ x + 2y–5 = 0(multiply by –1 on both the sides of the equation)



Question 21.

In an isosceles Δ PQR, PQ = PR. The base QR lies on the x–axis, P lies on the y– axis and 2x – 3y + 9 = 0 is the equation of PQ. Find the equation of the straight line along PR.


Answer:

Here it’s given that in an isosceles PQR, PQ = PR. The base QR lies on the x–axis, P lies on the y– axis and 2x – 3y + 9 = 0 is the equation of PQ.



The point P lies on the y–axis, so we have to put x = 0 in the equation for PQ


i.e.2x –3y + 9 = 0


⇒ 2(0)–3y + 9 = 0


⇒ –3y + 9 = 0


⇒ –3y = –9


⇒ 


Therefore, the point P is (0,3).


Now, to find the point Q which is on the x axis ,we have to substitute 0 in the place of y in the given equation QR.


2x –3y + 9 = 0


⇒ 2x–3(0) + 9 = 0


⇒ 2x + 9 = 0


⇒ 2x = –9


⇒ 


Therefore,the point Q becomes (


Now to find the equation of PR, where P is (0,3) and Q is (,we


have to use ,



⇒ 


⇒ 


⇒ 


⇒ –9y + 27 = 2(–3x)


⇒ –9y + 27 = –6x


⇒ –9y + 27 + 6x = 0


⇒ 2x–3y + 9 = 0(Divide by 3 on both the sides)




Exercise 5.6
Question 1.

The midpoint of the line joining (a,– b) and (3a, 5b) is
A. (–a, 2b)

B. (2a, 4b)

C. (2a, 2b)

D. (–a,– 3b)


Answer:

The midpoint of the line joining (a,– b) and (3a, 5b) is


,(mid point of line segment is ; where (x1,y1) and (x2,y2) are end points of the line segment)


i.e.


= (2a,2b)


Question 2.

The point P which divides the line segment joining the points A(1,– 3)and B(–3, 9) internally in the ratio 1:3 is
A. (2,1)

B. (0, 0)

C. 

D. (1, –2)


Answer:

Here P divides the line segment joining the points A(1,– 3)and B(–3, 9) internally in the ratio 1:3 .


Therefore coordinates of P are


;where k1 and k2 are the ratio in which the line is divided.


Now we substitute the values :



⇒ 


⇒ 


⇒ xp = 0,yp = 0


Question 3.

If the line segment joining the points A(3, 4) and B (14,– 3)meets the x–axis at P, then the ratio in which P divides the segment AB is
A. 4 : 3

B. 3 : 4

C. 2 : 3

D. 4 : 1


Answer:

Here the line segment joining the points A(3, 4) and


B (14,– 3)meets the x–axis at P.


Therefore coordinates of P are


;where k1 and k2 are the ratio in which the line is divided.


Now,



⇒  ,


(here yp = 0 the line meets the x–axis at P)


⇒ 


⇒ –3k1 + 4k2 = 0


⇒ –3k1 = –4k2


⇒ 


Question 4.

The centroid of the triangle with vertices at (–2,– 5), (–2,12) and (10, – 1)is
A. (6, 6)

B. (4, 4)

C. (3, 3)

D. (2, 2)


Answer:

Given the triangle with vertices at (–2,– 5),


(–2,12) and (10, – 1).


The centroid of the triangle ABC is 


∴ centroid of the given ∆ABC is


i.e.  = (2,2)


Question 5.

If (1, 2), (4, 6), (x, 6)and (3, 2)are the vertices of a parallelogram taken in order, then the value of x is
A. 6

B. 2

C. 1

D. 3


Answer:

Here we have the parallelogram with vertices


(1, 2), (4, 6), (x, 6)and (3, 2).


Let the vertices be A(1,2),B(4,6),C(x,6) and D(3,2)


Since the vertices are taken in order AC and BD are the diagonals of the parallelogram.


In a parallelogram diagonals bisect each other.


∴Mid–point of AC = Mid–point of BD


Mid–point of two points (x1,y1) and (x2,y2) is



Here Mid–point of AC = –––(1)


Mid–point of BD = –––(2)


(1) = (2)


⇒ 


Now we will equate the corresponding coordinates.


∴ 


⇒ 2(1 + x) = 7×2


⇒ 2 + 2x = 14


⇒ 2x = 12


⇒ x = 6


Question 6.

Area of the triangle formed by the points (0,0), (2, 0)and (0, 2)is
A. 1 sq. units

B. 2 sq. units

C. 4 sq. units

D. 8 sq. units


Answer:

Here we have the triangle with vertices (0,0), (2, 0)and (0, 2).


Area of the ∆ABC with vertices as A(x1,y1),B(x2,y2) and C(x3,y3) is



Let us say the vertices are A(0,0),B(2,0) and C(0,2).



⇒ 


⇒ 


⇒ 


⇒ Area = 2 sq.units


Hence Area is 2 sq.units.


Question 7.

Area of the quadrilateral formed by the points (1,1), (0,1), (0, 0)and (1, 0)is
A. 3 sq. units

B. 2 sq. units

C. 4 sq. units

D. 1 sq. units


Answer:

Here we have the quadrilateral formed by the points (1,1), (0,1), (0, 0)and (1, 0).


We have to calculate the Area of the quadrilateral with vertices as A(1,1), B(0,1), C(0, 0)and D(1, 0)


Let us divide the quadrilateral into 2 triangles,so Area of the quadrilateral will be sum of Areas of two triangles.


Let us say one ∆ is ABC and other ∆ is ADC


Now Area of ∆ABC is



When we substitute the values of the coordinates of the vertices


as A(1,1), B(0,1), C(0, 0),we get



⇒ 


Now Area of ∆ADC with vertices A(1,1), D(1, 0)


C(0, 0) is



⇒ 


Hence area is 1 sq.units.


Question 8.

The angle of inclination of a straight line parallel to x–axis is equal to
A. 0°

B. 60°

C. 45°

D. 90°


Answer:

Since the line is parallel to x–axis makes an angle of

0 degree with x–axis,the angle of inclination becomes 0°.


Question 9.

Slope of the line joining the points (3,– 2)and (–1, a) is , then the value of a is equal to
A. 1

B. 2

C. 3

D. 4


Answer:

Slope of the line joining the points (3,– 2)and (–1, a) is .


Slope of the line with end points as (x1,y1) and (x2,y2)


is .


∴ 


Given .


∴ 


⇒ 2(a + 2) = –3(–4)


⇒ 2a + 4 = 12


⇒ a + 2 = 6 (Divide by 2 on both the sides)


⇒ a = 6–2 = 4


∴ a = 4


Question 10.

Slope of the straight line which is perpendicular to the straight line joining the points (–2, 6)and (4, 8)is equal to
A. 

B. 3

C. –3

D. 


Answer:

The straight line is perpendicular to the straight line joining the points(–2, 6)and (4, 8).


The slope of line joining the points (–2, 6)and (4, 8) is m1.


i.e. 


Now the perpendicular line will have the slope m2 and as


the product of the slopes of 2 perpendicular lines is –1.


m1×m2 = –1


⇒ 


⇒ m2 = –3


Question 11.

The point of intersection of the straight lines 9x – y – 2 = 0 and 2x + y – 9 = 0 is
A. (–1, 7)

B. (7,1)

C. (1, 7)

D. (–1,– 7)


Answer:

We can get the point of intersection of the straight lines 9x – y – 2 = 0 and 2x + y – 9 = 0 by solving these equations.