Coordinate Geometry Class 9th Mathematics Term 1 Tamilnadu Board Solution

Class 9th Mathematics Term 1 Tamilnadu Board Solution
Exercise 5.1
  1. State whether the following statements are true / false. i. (5, 7) is a point in…
  2. Plot the following points in the coordinate system and specify their quadrant.…
  3. Write down the abscissa for the following points. i. (−7, 2) ii. (3, 5) iii. (8,…
  4. Write down the ordinate of the following points. i. (7, 5) ii. (2, 9) iii. (−5,…
  5. Plot the following points in the coordinate plane. i. (4, 2) ii. (4, −5) iii.…
  6. The ordinates of two points are each −6. How is the line joining them related…
  7. The abscissa of two points is 0. How is the line joining situated?…
  8. Mark the points A (2, 4), B (3, 4),C (3, 1) and D (2, 1) in the cartesian plane.…
  9. With rectangular axes plot the points O (0, 0), A (5, 0), B (5, 4). Find the…
  10. In a rectangle ABCD, the coordinates of A, B and D are (0, 0) (4, 0) (0, 3).…
Exercise 5.2
  1. (7, 8) and (−2, −3) Find the distance between the following pairs of points.…
  2. (6, 0) and (−2, 4) Find the distance between the following pairs of points.…
  3. (−3, 2) and (2, 0) Find the distance between the following pairs of points.…
  4. (−2, −8) and (−4, −6) Find the distance between the following pairs of points.…
  5. (−2, −3) and (3, 2) Find the distance between the following pairs of points.…
  6. (2, 2) and (3, 2) Find the distance between the following pairs of points.…
  7. (−2, 2) and (3, 2) Find the distance between the following pairs of points.…
  8. (7, 0) and (8, 0) Find the distance between the following pairs of points.…
  9. (0, 17) and (0, −1) Find the distance between the following pairs of points.…
  10. (5, 7) and the origin Find the distance between the following pairs of points.…
  11. (3, 7), (6, 5) and (15, −1) Show that the following points are collinear.…
  12. (3, −2), (−2, 8) and (0, 4) Show that the following points are collinear.…
  13. (1, 4), (3, −2) and (−1, 10) Show that the following points are collinear.…
  14. (6, 2), (2, −3) and (−2, −8) Show that the following points are collinear.…
  15. (4, 1), (5, −2) and (6, −5) Show that the following points are collinear.…
  16. (−2, 0), (4, 0) and (1, 3) Show that the following points form an isosceles…
  17. (1, −2), (−5, 1) and (1, 4) Show that the following points form an isosceles…
  18. (−1, −3), (2, −1) and (−1, 1) Show that the following points form an isosceles…
  19. (1, 3), (−3, -5) and (−3, 0) Show that the following points form an isosceles…
  20. (2, 3), (5, 7) and (1, 4) Show that the following points form an isosceles…
  21. (2, −3), (−6, −7) and (−8, −3) Show that the following points form a…
  22. (−11, 13), (−3, −1) and (4, 3) Show that the following points form a…
  23. (0, 0), (a, 0) and (0, b) Show that the following points form a right-angled…
  24. (10, 0), (18, 0) and (10, 15) Show that the following points form a…
  25. (5, 9), (5, 16) and (29, 9) Show that the following points form a right-angled…
  26. (0, 0), (10, 0) and (5, 5√3) Show that the following points form an equilateral…
  27. (a, 0), (−a, 0) and (0, a√3) Show that the following points form an equilateral…
  28. (2, 2), (−2, −2) and (−2√3, 2√3) Show that the following points form an…
  29. (√3, 2), (0,1) and (0, 3) Show that the following points form an equilateral…
  30. (−√3, 1), (2√3, −2) and (2√3, 4) Show that the following points form an…
  31. (−7, -5), (−4, 3), (5, 6) and (2, −2) Show that the following points taken in…
  32. (9, 5), (6, 0), (−2, −3) and (1, 2) Show that the following points taken in…
  33. (0, 0), (7, 3), (10, 6) and (3, 3) Show that the following points taken in…
  34. (−2, 5), (7, 1), (−2, −4) and (7, 0) Show that the following points taken in…
  35. (3, −5), (−5, −4), (7, 10) and (15, 9) Show that the following points taken in…
  36. (0, 0), (3, 4), (0, 8) and (−3, 4) Show that the following points taken in…
  37. (−4, −7), (−1, 2), (8, 5) and (5, −4) Show that the following points taken in…
  38. (1, 0), (5, 3), (2, 7) and (−2, 4) Show that the following points taken in…
  39. (2, −3), (6, 5), (−2, 1) and (−6, −7) Show that the following points taken in…
  40. (15, 20), (−3, 12), (−11, −6) and (7, 2) Show that the following points taken…
  41. (0, −1), (2, 1), (0, 3) and (−2, 1) Examine whether the following points taken…
  42. (5, 2), (1, 5), (−2, 1) and (2, −2) Examine whether the following points taken…
  43. (3, 2), (0, 5), (−3, 2) and (0, −1) Examine whether the following points taken…
  44. (12, 9), (20, −6), (5, −14) and (−3, 1) Examine whether the following points…
  45. (−1, 2), (1, 0), (3, 2) and (1, 4) Examine whether the following points taken…
  46. (8, 3), (0, −1), (−2, 3) and (6, 7) Examine whether the following points taken…
  47. (−1, 1), (0, 0), (3, 3) and (2, 4) Examine whether the following points taken…
  48. (−3, 0), (1, −2), (5, 6) and (1, 8) Examine whether the following points taken…
  49. If the distance between two points (x,7) and (1, 15) is 10, find x.…
  50. Show that (4, 1) is equidistant from the points (−10, 6) and (9, −13).…
  51. If two points (2, 3) and (−6, −5) are equidistant from the point (x, y), show…
  52. If the length of the line segment with end points (2, −6) and (2, y) is 4, find…
  53. Find the perimeter of the triangle with vertices (i) (0, 8), (6, 0) and origin;…
  54. Find the point on the y-axis equidistant from (−5, 2) and (9, −2) (Hint: A…
  55. Find the radius of the circle whose center is (3, 2) and passes through (−5,…
  56. Prove that the points (0, −5), (4, 3) and (−4, −3) lie on the circle centered…
  57. In the Fig. 5.20, PB is perpendicular segment from the point A (4, 3). If PA =…
  58. Find the area of the rhombus ABCD with vertices A (2, 0), B (5, -5), C (8, 0)…
  59. Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14)? Give reason.…
  60. If origin is the center of a circle with radius 17 units, find the coordinates…
  61. Show that (2, 1) is the circum-center of the triangle formed by the vertices…
  62. Show that the origin is the circum-center of the triangle formed by the…
  63. If the points A (6, 1), B (8, 2), C (9, 4) and D (p, 3) taken in order are the…
  64. The radius of the circle with center at the origin is 10 units. Write the…
Exercise 5.3
  1. The point (-2, 7) lies is the quadrantA. I B. II C. III D. IV
  2. The point (x, 0) where x 0 lies onA. OX B. OY C. OX’ D. OY’
  3. For a point A (a, b) lying in quadrant IIIA. a 0, b 0 B. a 0, b 0 C. a 0, b 0 D.…
  4. The diagonal of a square formed by the points (1, 0) (0, 1) (-1, 0) and (0, -1)…
  5. The triangle obtained by joining the points A (-5, 0) B (5, 0) and C (0, 6) isA.…
  6. The distance between the points (0, 8) and (0, -2) isA. 6 B. 100 C. 36 D. 10…
  7. (4, 1), (-2, 1), (7, 1) and (10, 1) are pointsA. on x-axis B. on a line parallel…
  8. The distance between the points (a, b) and (-a, -b) isA. 2a B. 2b C. 2a + 2b D.…
  9. The point which is on y-axis with ordinate -5 isA. (0, −5) B. (−5, 0) C. (5, 0)…
  10. The relation between p and q such that the point (p, q) is equidistant from (-4,…

Exercise 5.1
Question 1.

State whether the following statements are true / false.

i. (5, 7) is a point in the IV quadrant.

ii. (−2, −7) is a point in the III quadrant.

iii. (8, −7) lies below the x–axis.

iv. (5, 2) and (−7, 2) are points on the line parallel to y–axis.

v. (−5, 2) lies to the left of y–axis.

vi. (0, 3) is a point on x–axis.

vii. (−2, 3) lies in the II quadrant.

viii. (−10, 0) is a point on x–axis.

ix. (−2, −4) lies above x–axis.

x. For any point on the x–axis its y–coordinate is zero.


Answer:

i. (5,7) is point in the IV quadrant.


False


Reason: X –coordinate(abscissa) and y –coordinate (ordinate) both are positive. When both are positives, then they lie in the I quadrant.


ii. (–2, –7) is point in the III quadrant.


True


Reason: X–coordinate (Abscissa) and y –coordinate (ordinate) both are negative. When both are negatives, then they lie in the III quadrant.


iii. (8, −7) lies below the x–axis.


True


Reason: x – coordinate (Abscissa) is positive and y – coordinate (ordinate) is negative. Hence, this point lies in the IV quadrant. IV quadrant is the area below the x–axis.


iv. (5, 2) and (–7, 2) are points on the line parallel to y–axis.


False


Reason: (5, 2) and (–7, 2) are the line parallel to x–axis. Because, for any points to lie on line parallel to y–axis, the x–coordinates should be same. Hence, these points cannot lie on the line parallel to y–axis.


v. (–5, 2) lies to the left of y–axis.


True


Reason: x – coordinate (Abscissa) is negative and y – coordinate (ordinate) is positive. Hence, this point lies in the II quadrant. II quadrant is the area left of y–axis.


vi. (0, 3) is point on x–axis.


False


Reason: For any point on x–axis, the value of y–coordinate(ordinate) is 0. Hence, this point does not lie on x–axis.


vii. (–2, 3) lies in the II quadrant.


True


Reason: X – coordinate (Abscissa) is negative and y – coordinate (ordinate) is positive. Hence, this point lies in the II quadrant.


viii. (–10, 0) is point on x–axis.


True


Reason: For any point on the x–axis, the value of y–coordinate is zero. Hence, this point lies on the x–axis.


ix. (–2, –4) lies above x–axis


False


Reason: When both coordinates, i.e., x–coordinate and y–coordinate are negative, the point lies in the III quadrant. Therefore (–2, –4) lies in the III quadrant, which is below the axis.


x. For any point on the x–axis its y–coordinate is zero.


True



Question 2.

Plot the following points in the coordinate system and specify their quadrant.

i. (5, 2) ii. (−1, −1)

iii. (7, 0) iv. (−8, −1)

v. (0, −5) vi. (0, 3)

vii. (4, −5) viii. (0, 0)

ix. (1, 4) x. (−5, 7)


Answer:


i (5, 2) – I quadrant


ii (–1, –1) – III quadrant


iii (7, 0) – on X–axis


iv (–8, 1) – II quadrant


v (0, –5) – on down y–axis


vi (0, 3) – on y – axis


vii (4, –5) IV quadrant


viii (0, 0) – on origin


ix (1, 4) – I quadrant


x (–5, 7) – II quadrant



Question 3.

Write down the abscissa for the following points.

i. (−7, 2) ii. (3, 5)

iii. (8, −7) iv. (−5, −3)


Answer:

Abscissa is the x–coordinate of any point A (x, y)


i. (–7, 2)


Abscissa of point (–7, 2) is –7


ii. (3, 5)


Abscissa of point (3, 5) is 3


iii. (8, –7)


Abscissa of point (8, –7) is 8


iv. (–5, –3)


Abscissa of point (–5, –3) is –5



Question 4.

Write down the ordinate of the following points.

i. (7, 5) ii. (2, 9)

iii. (−5, 8) iv. (−7, −3)


Answer:

Ordinate is the y–coordinate of any point A (x, y)


i. (7, 5)


Ordinate of point (7, 5) is 5


ii. (2, 9)


Ordinate of point (2, 9) is 9


iii. (–5, 8)


Ordinate of point (–5, 8) is 8


iv. (–5, –3)


Ordinate of point (–5, –3) is –3



Question 5.

Plot the following points in the coordinate plane.

i. (4, 2) ii. (4, −5)

iii. (4, 0) iv. (4, −2)

How is the line joining them situated?


Answer:

Let (4, 2) be A, (4, –5) be B, (4,0) be C and (4, –2) be D.



The line joining the coordinates A, B, C and D is parallel to the y–axis.



Question 6.

The ordinates of two points are each −6. How is the line joining them related with reference to x–axis?


Answer:

Let the coordinates of two points i.e. A and B be (2, –6) and (–3, –6) respectively.



As we can see that, the line joining the point A and B is parallel to x–axis.



Question 7.

The abscissa of two points is 0. How is the line joining situated?


Answer:

Let the coordinate of two points i.e. A and B are (0, 3) and (0, –3) respectively.



As we can see that, the line joining the point A and B lies on the y–axis.



Question 8.

Mark the points A (2, 4), B (−3, 4),C (−3, −1) and D (2, −1) in the cartesian plane. State the figure obtained by joining A and B, B and C, C and D and D and A.


Answer:


To plot A (2, 4) move 2 units in positive x direction and 4 units in positive y direction.
To plot B (−3, 4) move 3 units in negative x direction and 4 units in positive y direction.
To plot C (−3, −1)move 3 units in negative x direction and 1 unit in negative y direction.
To plot D (2, −1)move 2 units in positive x direction and 1 unit in negative y direction.


Now use distance formula to find the lengths of each side,

For AB,

For AD,

For CD,

For BC,

Now AC,

For BD,

As AB = AC = BC = CD
Also AC = BD
Hence the given points make a square.


Question 9.

With rectangular axes plot the points O (0, 0), A (5, 0), B (5, 4). Find the coordinate of point C such that OABC forms a rectangle.


Answer:


For OABC to be square, the coordinate should be in a line where point B is and where it meets the y–axis. Therefore, the point C should be (0, 4).



Question 10.

In a rectangle ABCD, the coordinates of A, B and D are (0, 0) (4, 0) (0, 3). What are the coordinates of C?


Answer:

To obtain the coordinate C, extend a line from D towards right and extend a line from the coordinate B. the intersection point is the point C.



Hence, the coordinates of point C is (4, 3).




Exercise 5.2
Question 1.

Find the distance between the following pairs of points.

(7, 8) and (−2, −3)


Answer:

Formula used: 


(7, 8) and (–2, –3)


x1 = 7 and x2 = –2


y1 = 8 and y2 = –3


⇒ D = √ ((–2 – 7)2 + (–3 – 8)2)


⇒ D = √ ((–9)2 + (–11)2)


⇒ D = √ (81 + 121)


⇒ D = √ 202



Question 2.

Find the distance between the following pairs of points.

(6, 0) and (−2, 4)


Answer:

Formula used: 


(6, 0) and (–2, 4)


x1 = 6 and x2 = –2


y1 = 0 and y2 = 4


⇒ D = √ ((–2 – 6)2 + (4 – 0)2)


⇒ D = √ ((–8)2 + (4)2)


⇒ D = √ (64 + 16)


⇒ D = √ 80


⇒ D = √ (5 × 4 × 4)


⇒ D = 4√ 5



Question 3.

Find the distance between the following pairs of points.

(−3, 2) and (2, 0)


Answer:

Formula used: 


(–3, 2) and (2, 0)


x1 = –3 and x2 = 2


y1 = 2 and y2 = 0


⇒ D = √ ((2 – (–3)2 + (0 – 2)2)


⇒ D = √ ((2 + 3)2 + (0 – 2)2)


⇒ D = √ ((5)2 + (–2)2)


⇒ D = √ (25 + 4)


⇒ D = √ 29



Question 4.

Find the distance between the following pairs of points.

(−2, −8) and (−4, −6)


Answer:

Formula used: 


(–2, –8) and (–4, –6)


x1 = –2 and x2 = –4


y1 = –8 and y2 = –6


⇒ D = √ ((–4 – (–2))2 + (–6 – (–8))2)


⇒ D = √ ((–4 + 2)2 + (–6 + 8)2)


⇒ D = √ ((–2)2 + (2)2)


⇒ D = √ (4 + 4)


⇒ D = √ 8


⇒ D = √ (2 × 2 × 2)


⇒ D = 2√ 2



Question 5.

Find the distance between the following pairs of points.

(−2, −3) and (3, 2)


Answer:

Formula used: 


(–2, –3) and (3, 2)


x1 = –2 and x2 = 3


y1 = –3 and y2 = 2


⇒ D = √ ((3 – (–2))2 + (2 – (–3))2)


⇒ D = √ ((3 + 2)2 + (2 + 3)2)


⇒ D = √ ((5)2 + (5)2)


⇒ D = √ (25 + 25)


⇒ D = √ 50


⇒ D = √ (5 × 5 × 2)


⇒ D = 5√ 2



Question 6.

Find the distance between the following pairs of points.

(2, 2) and (3, 2)


Answer:

Formula used: 


(2, 2) and (3, 2)


x1 = 2 and x2 = 3


y1 = 2 and y2 = 2


⇒ D = √ ((3 – 2)2 + (2 – 2)2)


⇒ D = √ ((1)2 + (0)2)


⇒ D = √ (1 + 0)


⇒ D = √ 1


⇒ D = 1



Question 7.

Find the distance between the following pairs of points.

(−2, 2) and (3, 2)


Answer:

Formula used: 


(–2, 2) and (3, 2)


x1 = –2 and x2 = 3


y1 = 2 and y2 = 2


⇒ D = √ ((3 – (–2))2 + (2 – 2)2)


⇒ D = √ ((5)2 + (0)2)


⇒ D = √ (25 + 0)


⇒ D = √ 25


⇒ D = √ (5 × 5)


⇒ D = 5



Question 8.

Find the distance between the following pairs of points.

(7, 0) and (8, 0)


Answer:

Formula used: 


(7, 0) and (–8, 0)


x1 = 7 and x2 = –8


y1 = 0 and y2 = 0


⇒ D = √ ((–8 – 7)2 + (0 – 0)2)


⇒ D = √ ((–15)2 + (0)2)


⇒ D = √ (225 + 0)


⇒ D = √ 225


⇒ D = √ (5 × 3 × 5 × 5)


⇒ D = 5 × 3


⇒ D = 15



Question 9.

Find the distance between the following pairs of points.

(0, 17) and (0, −1)


Answer:

Formula used: 


(0, 17) and (0, –1)


x1 = 0 and x2 = 0


y1 = 17 and y2 = –1


⇒ D = √ ((0 – 0)2 + (–1 – 17)2)


⇒ D = √ ((0)2 + (–18)2)


⇒ D = √ (0 + 324)


⇒ D = √ 324


⇒ D = √ (18 × 18)


⇒ D = 18



Question 10.

Find the distance between the following pairs of points.

(5, 7) and the origin


Answer:

Formula used: 


(5, 7) and (0, 0)


x1 = 5 and x2 = 0


y1 = 7 and y2 = 0


⇒ D = √ ((0 – 5)2 + (0 – 7)2)


⇒ D = √ ((–5)2 + (–7)2)


⇒ D = √ (25 + 49)


⇒ D = √ 74



Question 11.

Show that the following points are collinear.

(3, 7), (6, 5) and (15, −1)


Answer:

Formula used: 


(3, 7), (6, 5) and (15, –1)


Let the points be A (15, –1), B (6, 5) and C (3, 7)


Distance of AB


⇒ AB = √ (6 – 15)2 + (5 – (–1))2


⇒ AB = √ (–9)2 + (6)2


⇒ AB = √ (81 + 36)


⇒ AB = √ 117 = √ 3 × 3 × 13


⇒ AB = 3√13


Distance of BC


⇒ BC = √ (3 – 6)2 + (7 – 5)2


⇒ BC= √ (3)2 + (2)2


⇒ BC = √ (9 + 4)


⇒ BC= √ 13


Distance of AC


⇒ AC = √ (3 – 15)2 + (7 – (–1))2


⇒ AC = √ (3 – 15)2 + (7 + 1)2


⇒ AC= √ (–12)2 + (8)2


⇒ AC = √ (144 + 64)


⇒ AC= √ 208 = √ 4 × 4 × 13


⇒ AC = 4√13


i.e. AB + BC = AC


⇒ 3√13 + √13 = 4√13


∴ A, B and C are collinear



Question 12.

Show that the following points are collinear.

(3, −2), (−2, 8) and (0, 4)


Answer:

Formula used: 


(3, 2), (–2, 8) and (0, 4)


Let A (–2, 8), B (0, 4) and C (3, 2)


Distance of AB


⇒ AB = √ ((0 – (–2))2 + (4 – 8)2)


⇒ AB = √ (2)2 + (–4)2


⇒ AB = √ (4 + 16)


⇒ AB = √20


Distance of BC


⇒ BC = √ ((3 – 0)2 + (2 – 4)2)


⇒ BC = √ (3)2 + (–2)2


⇒ BC = √ (9 + 4)


⇒ BC = √13


Distance of AC


⇒ AC = √ ((3 – (–2))2 + (2 – 8)2)


⇒ AC = √ (5)2 + (–6)2


⇒ AC = √ (25 + 36)


⇒ AC = √ 61



Question 13.

Show that the following points are collinear.

(1, 4), (3, −2) and (−1, 10)


Answer:

Formula used: 


(1, 4), (3, –2) and (–1, 10)


Let A (–1, 10), B (1, 4) and C (3, –2)


Distance of AB


⇒ AB =√ ((1 – (–1))2 + (4 – 10)2)


⇒ AB = √ ((1 + 1)2 + (4 – 10)2)


⇒ AB = √ (2)2 + (–6)2


⇒ AB = √ (4 + 36)


⇒ AB = √ 40


Distance of BC


⇒ BC =√ ((3 – 1)2 + (–2 – 4)2)


⇒ BC = √ (2)2 + (–6)2


⇒ BC = √ (4 + 36)


⇒ BC = √ 40


Distance of AC


⇒ AC =√ ((3 – (–1))2 + (–2 – 10)2)


⇒ AC = √ ((3 + 1)2 + (2 – 10)2)


⇒ AC = √ (4)2 + (–8)2


⇒ AC = √ (16 + 64)


⇒ AC = √ 80


i.e. AB + BC = AC


⇒ √40 + √40 = √80


∴ A, B and C are collinear.



Question 14.

Show that the following points are collinear.

(6, 2), (2, −3) and (−2, −8)


Answer:

Formula used: 


(6, 2), (2, –3) and (–2, –8)


Let A (6, 2), B (2, –3) and C (–2, –8)


Distance of AB


⇒ AB =√ ((2 – (6))2 + (–3 – 2)2)


⇒ AB = √ (4)2 + (–5)2


⇒ AB = √ (16 + 25)


⇒ AB = √ 41


Distance of BC


⇒ BC =√ ((–2 – 2)2 + (–8 – (–3))2)


⇒ BC = √ ((–2 – 2)2 + (–8 + 3)2)


⇒ BC = √ (–4)2 + (–5)2


⇒ BC = √ (16 + 25)


⇒ BC = √ 41


Distance of AC


⇒ AC =√ ((–2 – 6)2 + (–8 – 2)2)


⇒ AC = √ (–8)2 + (–10)2


⇒ AC = √ (64 + 100)


⇒ AC = √ 164 = √ 2 × 2 × 41


⇒ AC =2√ 41


i.e. AB + BC = AC


⇒ √41 + √41 = 2√41


∴ A, B and C are collinear.



Question 15.

Show that the following points are collinear.

(4, 1), (5, −2) and (6, −5)


Answer:

Formula used: 


(4, 1), (5, –2) and (6, –5)


Let A (4, 1), B (5, –2) and C (6, –5)


Distance of AB


⇒ AB =√ ((5 – 4)2 + (–2 – 1)2)


⇒ AB = √ (1)2 + (–3)2


⇒ AB = √ (1 + 9)


⇒ AB = √10


Distance of BC


⇒ BC =√ ((6 – 5)2 + (–5 – (–2))2)


⇒ BC = √ (6 – 5)2 + (–5 + 2)2)


⇒ BC = √ (1)2 + (–3)2


⇒ BC = √ (1 + 9)


⇒ BC = √ 10


Distance of AC


⇒ AC =√ ((6 – 4)2 + (–5 – 1)2)


⇒ AC = √ (2)2 + (–6)2


⇒ AC = √ (4 + 36)


⇒ AC = √20 =


i.e. AB + BC = AC


⇒ √10 + √10 = √20


Squaring both sides


⇒ (√10)2 + (√10)2 = (√20)2


⇒ 10 + 10 = 20


∴ A, B and C are collinear.



Question 16.

Show that the following points form an isosceles triangle.

(−2, 0), (4, 0) and (1, 3)


Answer:

Formula used: 


(–2, 0), (4,0) and (1, 3)


Let the point be A (1, 3) B (–2, 0) and C (4, 0)


Distance of AB


⇒ AB = √ ((–2 – 1)2 + (0 – 3)2)


⇒ AB = √ ((–3)2 + (–3)2)


⇒ AB = √ (9 + 9)


⇒ AB = √ 18 = 3√ 2


Distance of AC


⇒ AC = √ ((4 – 1)2 + (0 – 3)2)


⇒ AC = √ ((3)2 + (–3)2)


⇒ AC = √ (9 + 9)


⇒ AC = √ 18 = 3√2


Distance of BC


⇒ BC = √ ((4 – (–2))2 + (0 – 0)2)


⇒ BC = √ ((6)2 + (0)2)


⇒ BC = √ (36 + 0)


⇒ BC = √ 36 = 6


We notice that AB = AC =3√2


∴ Points A, B and C are coordinates of an isosceles triangle.



Question 17.

Show that the following points form an isosceles triangle.

(1, −2), (−5, 1) and (1, 4)


Answer:

Formula used: 


(1, −2), (−5, 1) and (1, 4)


Let the point be A (–5, 1) B (1, –2) and C (1, 4)


Distance of AB


⇒ AB = √ (1 – (–5))2 + (–2 – 1)2)


⇒ AB = √ (1 + 5)2 + (–2 – 1)2


⇒ AB = √ ((6)2 + (–3)2)


⇒ AB = √ (36 + 9)


⇒ AB = √ 45 = 3√5


Distance of AC


⇒ AC = √ ((1 – (–5))2 + (4 – 1)2)


⇒ AC = √ ((1 + 5)2 + (4 – 1)2)


⇒ AC = √ ((6)2 + (3)2)


⇒ AC = √ (36 + 9)


⇒ AC = √45 = 3√5


Distance of BC


⇒ BC = √ ((1 – 1)2 + (4 – (–2)2)


⇒ BC = √ ((1 – 1)2 + (4 + 22)


⇒ BC = √ ((0)2 + (6)2)


⇒ BC = √ (0 + 36)


⇒ BC = √ 36 = 6


We notice that AB = AC =3√5


∴ Points A, B and C are coordinates of an isosceles triangle.



Question 18.

Show that the following points form an isosceles triangle.

(−1, −3), (2, −1) and (−1, 1)


Answer:

Formula used: 


(−1, −3), (2, −1) and (−1, 1)


Let the point be A (2, –1) B (–1, –3) and C (–1, 1)


Distance of AB


⇒ AB = √ ((–1 – 2)2 + (–3 – (–1))2)


⇒ AB = √ ((–1 – 2)2 + (–3 + 1)2)


⇒ AB = √ ((–3)2 + (–2)2)


⇒ AB = √ (9 + 4)


⇒ AB = √ 13


Distance of AC


 AC = √ ((–1 – 2)2 + (1 – (–1))2)


⇒ AC = √ ((–1 – 2)2 + (1 + 1)2)


⇒ AC = √ ((–3)2 + (2)2)


⇒ AC = √ (9 + 4)


⇒ AC = √ 13


Distance of BC


⇒ BC = √ ((–1 – (–1))2 + (1 – (–3))2)


⇒ BC = √ ((–1 + 1))2 + (1 + 3)2)


⇒ BC = √ ((0)2 + (4)2)


⇒ BC = √ (0 + 16)


⇒ BC = √ 16


We notice that AB = AC = √13


∴ Points A, B and C are coordinates of an isosceles triangle.



Question 19.

Show that the following points form an isosceles triangle.

(1, 3), (−3, –5) and (−3, 0)


Answer:

Formula used: 


(1, 3), (–3, –5) and (–3, 0)


Let the point be A (–3, 0) B (1, 3) and C (–3, –5)


Distance of AB


⇒ AB = √ ((1 – (–3))2 + (3 – 0)2)


⇒ AB = √ ((1 + 3)2 + (3 – 0)2)


⇒ AB = √ ((4)2 + (3)2)


⇒ AB = √ (16 + 9)


⇒ AB = √25 = 5


Distance of AC


⇒ AC = √ ((–3 – (–3))2 + (–5 – 0)2)


⇒ AC = √ ((–3 + 3)2 + (–5 + 0)2)


⇒ AC = √ ((0)2 + (–5)2)


⇒ AC = √ (0 + 25)


⇒ AC = √25 = 5


Distance of BC


⇒ BC = √ ((–3 – 1)2 + (–5 – 3)2)


⇒ BC = √ ((–4)2 + (–8)2)


⇒ BC = √ (16 + 64)


⇒ BC = √ 80


We notice that AB = AC = 5


∴ Points A, B and C are coordinates of an isosceles triangle.



Question 20.

Show that the following points form an isosceles triangle.

(2, 3), (5, 7) and (1, 4)


Answer:

Formula used: 


(2, 3), (5, 7) and (1, 4)


Let the point be A (5, 7) B (2, 3) and C (1, 4)


Distance of AB


⇒ AB = √ (2 – 5)2 + (3 – 7)2)


⇒ AB = √ ((–3)2 + (–4)2)


⇒ AB = √ (9 + 16)


⇒ AB = √ 25 = 5


Distance of AC


⇒ AC = √ ((1 – 5)2 + (4 – 7)2)


⇒ AC = √ ((–4)2 + (–3)2)


⇒ AC = √ (16 + 9)


⇒ AC = √ 25 = 5


Distance of BC


⇒ BC = √ ((1 – 2)2 + (4 – 3)2)


⇒ BC = √ ((–1)2 + (1)2)


⇒ BC = √ (1 + 1)


⇒ BC = √ 2


We notice that AB = AC = 5


∴ Points A, B and C are coordinates of an isosceles triangle.



Question 21.

Show that the following points form a right–angled triangle.

(2, −3), (−6, −7) and (−8, −3)


Answer:

Formula used: 


(2, –3), (–6, –7) and (–8, –3)


Let the points be A (2, –3), B (–6, –7) and C (–8, –3)


Distance of AB


⇒ AB = √ ((–6 – 2)2 + (–7 – (–3))2)


⇒ AB = √ ((–6 – 2)2 + (–7 + 3)2)


⇒ AB = √ ((–8)2 + (–4)2)


⇒ AB = √ (64 + 16)


⇒ AB = √ 80


Distance of BC


⇒ B C= √ ((–8 – (–6))2 + (–3 – (–7))2)


⇒ BC = √ ((–8 + 6)2 + (–3 + 7)2)


⇒ BC = √ ((–2)2 + (4)2)


⇒ BC = √ (4 + 16)


⇒ BC = √ 20


Distance of AC


⇒ AC = √ ((–8 – 2)2 + (–3 – (–3))2)


⇒ AC = √ ((–8 – 2)2 + (–3 + 3)2)


⇒ AC = √ ((–10)2 + (0)2)


⇒ AC = √ (100 + 0)


⇒ AC = √ 100


i.e. AB2 + BC2


= (√80)2 + (√20)2


= 80 + 20


= 100 = (AC)2


Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.



Question 22.

Show that the following points form a right–angled triangle.

(−11, 13), (−3, −1) and (4, 3)


Answer:

Formula used: 


(–11, 13), (–3, –1) and (4, 3)


Let the points be A (–11, 13), B (–3, –1) and C (4, 3)


Distance of AB


⇒ AB = √ ((–3 – (–11))2 + (–1 – 13)2)


⇒ AB = √ ((–3 + 11)2 + (–1 – 13)2)


⇒ AB = √ ((8)2 + (–14)2)


⇒ AB = √ (64 + 196)


⇒ AB = √260


Distance of BC


⇒ B C= √ ((4 – (–3))2 + (3 – (–1))2)


⇒ BC = √ ((4 + 3)2 + (3 + 1)2)


⇒ BC = √ ((7)2 + (4)2)


⇒ BC = √ (49 + 16)


⇒ BC = √ 65


Distance of AC


⇒ AC = √ ((4 – (–11))2 + (3 – 13))2)


⇒ AC = √ ((4 + 11)2 + (3 – 13)2)


⇒ AC = √ ((15)2 + (–10)2)


⇒ AC = √ (225 + 100)


⇒ AC = √ 325


i.e. AB2 + BC2


= (√260)2 + (√65)2


= 260 + 65


= 325 = (AC)2


Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.



Question 23.

Show that the following points form a right–angled triangle.

(0, 0), (a, 0) and (0, b)


Answer:

Formula used: 


(0, 0), (a, 0) and (0, b)


Let the points be A (0, 0), B (a, 0) and C (0, b)


Distance of AB


⇒ AB = √ ((a – 0)2 + (0 – 0)2)


⇒ AB = √ ((a)2 + (0)2


⇒ AB = √ a2


Distance of BC


⇒ BC = √ ((0 – a)2 + (b – 0)2)


⇒ BC = √ ((–a)2 + (b)2


⇒ BC = √ a2 + b2


Distance of AC


⇒ AC = √ ((0 – 0)2 + (b – 0)2)


⇒ AC = √ ((0)2 + (b)2


⇒ AC = √ b2


i.e. AB2 + AC2


= (√a2)2 + (√b2)2


= a2 + b2 = BC2


Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.



Question 24.

Show that the following points form a right–angled triangle.

(10, 0), (18, 0) and (10, 15)


Answer:

Formula used: 


(10, 0), (18, 0) and (10, 15)


Let the points be A (10, 15), B (10, 0) and C (18, 0)


Distance of AB


⇒ AB = √ ((10 – 10))2 + (0 – 15)2)


⇒ AB = √ ((0)2 + (–15)2)


⇒ AB = √ (0 + 225)


⇒ AB = √225


Distance of BC


⇒ B C= √ ((18 – 10)2 + (0 – 0)2)


⇒ BC = √ ((8)2 + (0)2)


⇒ BC = √ (64 + 0)


⇒ BC = √ 64


Distance of AC


⇒ AC = √ ((18 – 10)2 + (0 – 15))2)


⇒ AC = √ ((8)2 + (–15)2)


⇒ AC = √ (64 + 225)


⇒ AC = √289


i.e. AB2 + BC2


= (√225)2 + (√64)2


= 225 + 64


= 289 = (AC)2


Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.



Question 25.

Show that the following points form a right–angled triangle.

(5, 9), (5, 16) and (29, 9)


Answer:

Formula used: 


(5, 9), (5, 16) and (29, 9)


Let the points be A (5, 16), B (5, 9) and C (29, 9)


Distance of AB


⇒ AB = √ ((5 – 5)2 + (9 – 16)2)


⇒ AB = √ ((0)2 + (–7)2)


⇒ AB = √ (0 + 49)


⇒ AB = √49


Distance of BC


⇒ B C= √ ((29 – 5)2 + (9 – 9)2)


⇒ BC = √ ((24)2 + (0)2)


⇒ BC = √ (576 + 0)


⇒ BC = √576


Distance of AC


⇒ AC = √ ((29 – 5)2 + (9 – 16))2)


⇒ AC = √ ((24)2 + (–7)2)


⇒ AC = √ (576 + 49)


⇒ AC = √ 625


i.e. AB2 + BC2


= (√49)2 + (√576)2


= 49 + 576


= 625 = (AC)2


Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.



Question 26.

Show that the following points form an equilateral triangle.

(0, 0), (10, 0) and (5, 5√3)


Answer:

Formula used: 


(0, 0), (10, 0) and (5, 5√3)


Let the points be A (0, 0), B (10, 0) and C (5, 5√3)


Distance of AB


⇒ AB = √ ((10 – 0)2 + (0 – 0)2)


⇒ AB = √ ((10)2 + (0)2)


⇒ AB = √ (100 + 0)


⇒ AB = √100


⇒ AB = 10


Distance of BC


⇒ B C= √ ((5 – 10)2 + (5√3 – 0)2)


⇒ BC = √ ((–5)2 + (5√3)2)


⇒ BC = √ (25 + 75)


⇒ BC = √100


⇒ BC = 10


Distance of AC


⇒ AC = √ ((5 – 0)2 + (5√3 – 0))2)


⇒ AC = √ ((5)2 + (5√3)2)


⇒ AC = √ (25 + 75)


⇒ AC = √ 100


⇒ AC = 10


∴ AB = BC = AC = 10


Since, all the sides are equal the points form an equilateral triangle.



Question 27.

Show that the following points form an equilateral triangle.

(a, 0), (−a, 0) and (0, a√3)


Answer:

Formula used: 


(a, 0), (–a, 0) and (0, a√3)


Let the points be A (a, 0), B (–a, 0) and C (0, a√3)


Distance of AB


⇒ AB = √ ((–a – a)2 + (0 – 0)2)


⇒ AB = √ ((–2a)2 + (0)2)


⇒ AB = √ (4a2 + 0)


⇒ AB = √4a2


⇒ AB = 2a


Distance of BC


⇒ B C= √ ((0 – a)2 + (a√3 – 0)2)


⇒ BC = √ ((–a)2 + (a√3)2)


⇒ BC = √ (a2 + 3a2)


⇒ BC = √4a2


⇒ BC = 2a


Distance of AC


⇒ AC = √ ((0 – a)2 + (a√3 – 0))2)


⇒ AC = √ ((–a)2 + (a√3)2)


⇒ AC = √ (a2 + 3a2)


⇒ AC = √ 4a2


⇒ AC = 2a


∴ AB = BC = AC = 2a


Since, all the sides are equal the points form an equilateral triangle.



Question 28.

Show that the following points form an equilateral triangle.

(2, 2), (−2, −2) and (−2√3, 2√3)


Answer:

Formula used: 


(2, 2), (–2, –2) and (–2√3, 2√3)


Let the points be A (2, 2), B (–2, –2) and C (–2√3, 2√3)


Distance of AB


⇒ AB = √ ((–2 – 2)2 + (–2 – 2)2)


⇒ AB = √ ((–4)2 + (–4)2)


⇒ AB = √ (16 + 16)


⇒ AB = √32


⇒ AB = 4√2


Distance of BC


⇒ B C= √ ((–2√3 – (–2))2 + (2√3 – (–2))2)


⇒ B C= √ ((–2√3 + 2))2 + (2√3 + 2)2)


⇒ BC = √ (((–2√3)2 + 2 (–2√3) (2) + (2)2) + ((2√3)2 + 2 (2√3) (2) + (2)2))


⇒ BC = √ (12 – 8√3 + 4 + 12 + 8√3 + 4)


⇒ BC = √ (12 + 4 + 12 + 4


⇒ BC = √ 32


⇒ BC = 4√2


Distance of AC


⇒ AC= √ ((–2√3 – 2))2 + (2√3 – 2)2)


⇒ AC = √ (((–2√3)2 + 2 (–2√3) (–2) + (2)2) + ((2√3)2 + 2 (2√3) (–2) + (–2)2))


⇒ AC = √ (12 + 8√3 + 4 + 12 – 8√3 + 4)


⇒ AC = √ (12 + 4 + 12 + 4)


⇒ AC = √ 32


⇒ AC = 4√2


∴ AB = BC = AC = 4√2


Since, all the sides are equal the points form an equilateral triangle.



Question 29.

Show that the following points form an equilateral triangle.

(√3, 2), (0,1) and (0, 3)


Answer:

Formula used: 


(√3, 2), (0, 1) and (0, 3)


Let the points be A (√3, 2), B (0, 1) and C (0, 3)


Distance of AB


⇒ AB = √ ((0 – √3)2 + (1 – 2)2)


⇒ AB = √ ((√3)2 + (–1)2)


⇒ AB = √ (3 + 1)


⇒ AB = √4


⇒ AB = 2


Distance of BC


⇒ B C= √ ((0 – 0)2 + (3 – 1)2)


⇒ BC = √ ((0)2 + (2)2)


⇒ BC = √ (0 + 4)


⇒ BC = √4


⇒ BC = 2


Distance of AC


⇒ AC = √ ((0 – √3)2 + (3 – 2))2)


⇒ AC = √ ((√3)2 + (1)2)


⇒ AC = √ (3 + 1)


⇒ AC = √ 4


⇒ AC = 2


∴ AB = BC = AC = 2


Since, all the sides are equal the points form an equilateral triangle.



Question 30.

Show that the following points form an equilateral triangle.

(−√3, 1), (2√3, −2) and (2√3, 4)


Answer:

Formula used: 


(–√3, 1), (2√3, –2) and (2√3, 4)


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


Distance of AB


⇒ AB = √ ((2√3 – (–√3))2 + (–2 – 1)2)


⇒ AB = √ ((2√3 + √3))2 + (–2 – 1)2)


⇒ AB = √ ((12 + 12 + 3)2 + (–3)2)


⇒ AB = √ (27 + 9)


⇒ AB = √36


⇒ AB = 6


Distance of BC


⇒ B C= √ ((2√3 – 2√3)2 + (4 – (–2))2)


⇒ B C= √ ((2√3 – 2√3)2 + (4 + 2)2)


⇒ BC = √ ((0)2 + (6)2)


⇒ BC = √ (0 + 36)


⇒ BC = √36


⇒ BC = 6


Distance of AC


⇒ AC = √ ((2√3 – (–√3))2 + (4 – 1))2)


⇒ AC = √ ((2√3 + √3))2 + (4 – 1))2)


⇒ AC = √ ((3√3)2 + (3)2)


⇒ AC = √ (27 + 9)


⇒ AC = √ 36


⇒ AC = 6


∴ AB = BC = AC = 6


Since, all the sides are equal the points form an equilateral triangle.



Question 31.

Show that the following points taken in order form the vertices of a parallelogram.

(−7, –5), (−4, 3), (5, 6) and (2, −2)


Answer:

Formula used: 


(–7, –5), (–4, 3), (5, 6) and (2, –2)


Let A, B, C and D represent the points (–7, –5), (–4, 3), (5, 6) and (2, –2)


Distance of AB


⇒ AB = √ ((–4 – (–7)))2 + (3 – (–5))2)


⇒ AB = √ ((–4 + 7))2 + (3 + 5)2)


⇒ AB = √ ((3)2 + (8)2)


⇒ AB = √ (9 + 64)


⇒ AB = √73


Distance of BC


⇒ BC= √ ((5 – (–4))2 + (6 – 3)2)


⇒ BC= √ ((5 + 4))2 + (6 – 3)2)


⇒ BC = √ ((9)2 + (3)2)


⇒ BC = √ (81 + 9)


⇒ BC = √ 90


Distance of CD


⇒ CD = √ ((2 – 5)2 + (–2 – 6)2)


⇒ CD = √ ((–3)2 + (–8)2)


⇒ CD = √ (9 + 64)


⇒ CD = √73


Distance of AD


⇒ AD = √ ((2 – (–7)))2 + (–2 – (–5))2)


⇒ AD = √ ((2 + 7))2 + (–2 + 5)2)


⇒ AD = √ ((9)2 + (3)2)


⇒ AD = √ (81 + 9)


⇒ AD = √ 90


So, AB = CD = √73 and BC = AD = √90


i.e., The opposite sides are equal. Hence ABCD is a parallelogram.



Question 32.

Show that the following points taken in order form the vertices of a parallelogram.

(9, 5), (6, 0), (−2, −3) and (1, 2)


Answer:

Formula used: 


(9,5), (6, 0), (–2, –3) and (1, 2)


Let A, B, C and D represent the points (9, 5), (6, 0), (–2, –3) and (1, 2)


Distance of AB


⇒ AB = √ ((6 – 9))2 + (0 – 5)2)


⇒ AB = √ ((–3)2 + (5)2)


⇒ AB = √ (9 + 25)


⇒ AB = √ 34


Distance of BC


⇒ BC= √ ((–2 – 6)2 + (–3 – 0)2)


⇒ BC = √ ((–8)2 + (–3)2)


⇒ BC = √ (64 + 9)


⇒ BC = √73


Distance of CD


⇒ CD = √ ((1 – (–2))2 + (2 – (–3))2)


⇒ CD = √ ((1 + 2)2 + (2 + 3))2)


⇒ CD = √ ((3)2 + (5)2)


⇒ CD = √ (9 + 25)


⇒ CD = √36


Distance of AD


⇒ AD = √ ((1 – 9))2 + (2 – 5)2)


⇒ AD = √ ((–8)2 + (–3)2)


⇒ AD = √ (64 + 9)


⇒ AD = √ 73


So, AB = CD = √36 and BC = AD = √73


i.e., The opposite sides are equal. Hence ABCD is a parallelogram.



Question 33.

Show that the following points taken in order form the vertices of a parallelogram.

(0, 0), (7, 3), (10, 6) and (3, 3)


Answer:

Formula used: 


(0,0) (7, 3), (10, 6) and (3, 3)


Let A, B, C and D represent the points (0, 0), (7, 3), (10, 6) and (3, 3)


Distance of AB


⇒ AB = √ ((7 – 0))2 + (3 – 0)2)


⇒ AB = √ ((7)2 + (3)2)


⇒ AB = √ (49 + 9)


⇒ AB = √ 58


Distance of BC


⇒ BC= √ ((10 – 7)2 + (6 – 3)2)


⇒ BC = √ ((3)2 + (3)2)


⇒ BC = √ (9 + 9)


⇒ BC = √18


Distance of CD


⇒ CD = √ ((3 – 10)2 + (3 – 6)2)


⇒ CD = √ ((–7)2 + (–3)2)


⇒ CD = √ (49 + 9)


⇒ CD = √58


Distance of AD


⇒ AD = √ ((3 – 0))2 + (3 – 0)2)


⇒ AD = √ ((3)2 + (3)2)


⇒ AD = √ (9 + 9)


⇒ AD = √18


So, AB = CD = √58 and BC = AD = √18


i.e., The opposite sides are equal. Hence ABCD is a parallelogram.



Question 34.

Show that the following points taken in order form the vertices of a parallelogram.

(−2, 5), (7, 1), (−2, −4) and (7, 0)


Answer:

Formula used: 


(–2, 5), (7, 1), (–2, –4) and (7, 0)


Let A, B, C and D represent the points (–2, 5), (7, 1), (–2, –4) and (7, 0)


Distance of AB


⇒ AB = √ ((7 – (–2)))2 + (1 – 5)2)


⇒ AB = √ ((7 + 2))2 + (1 – 5)2)


⇒ AB = √ ((9)2 + (–4)2)


⇒ AB = √ (81 + 16)


⇒ AB = √ 97


Distance of BC


⇒ BC= √ ((–2 – 7)2 + (–4 – 1)2)


⇒ BC = √ ((–9)2 + (–5)2)


⇒ BC = √ (81 + 25)


⇒ BC = √106


Distance of CD


⇒ CD = √ ((7 – (–2))2 + (0 – (–4))2)


⇒ CD = √ ((7 + 2)2 + (0 + 4))2)


⇒ CD = √ ((9)2 + (4)2)


⇒ CD = √ (81 + 16)


⇒ CD = √97


Distance of AD


⇒ AD = √ ((7 – (–2))2 + (0 – 5)2)


⇒ AD = √ ((7 + 2)2 + (0 – 5)2)


⇒ AD = √ ((9)2 + (–5)2)


⇒ AD = √ (81 + 25)


⇒ AD = √ 106


So, AB = CD = √97 and BC = AD = √106


i.e., The opposite sides are equal. Hence ABCD is a parallelogram.



Question 35.

Show that the following points taken in order form the vertices of a parallelogram.

(3, −5), (−5, −4), (7, 10) and (15, 9)


Answer:

Formula used: 


(3, –5), (–5, –4), (7, 10) and (15, 9)


Let A, B, C and D represent the points (3, –5), (–5, –4), (7, 10) and (15, 9)


Distance of AB


⇒ AB = √ ((–5 – 3)2 + ((–4 – (–5))2)


⇒ AB = √ ((–5 – 3))2 + (–4 + 5)2)


⇒ AB = √ ((–8)2 + (1)2)


⇒ AB = √ (64 + 1)


⇒ AB = √ 65


Distance of BC


⇒ BC= √ ((7 – (–5))2 + (10 – (–4))2)


⇒ BC= √ ((7 + 5)2 + (10 + 4)2)


⇒ BC = √ ((12)2 + (14)2)


⇒ BC = √ (144 + 196)


⇒ BC = √ 340


Distance of CD


⇒ CD = √ ((15 – 7)2 + (9 – 10)2)


⇒ CD = √ ((8)2 + (–1)2)


⇒ CD = √ (64 + 1)


⇒ CD = √65


Distance of AD


⇒ AD = √ ((15 – 3)2 + (9 – (–5))2)


⇒ AD = √ ((15 – 3)2 + (9 + 5)2)


⇒ AD = √ ((12)2 + (14)2)


⇒ AD = √ (144 + 196)


⇒ AD = √ 340


So, AB = CD = √65 and BC = AD = √340


i.e., The opposite sides are equal. Hence ABCD is a parallelogram.



Question 36.

Show that the following points taken in order form the vertices of a rhombus.

(0, 0), (3, 4), (0, 8) and (−3, 4)


Answer:

Formula used: 


(0, 0), (3, 4), (0, 8) and (–3, 4)


Let the vertices be taken as A (0, 0), B (3, 4), C (0, 8) and D (–3, 4).


Distance of AB


⇒ AB = √ ((3 – 0)2 + ((4 – 0)2)


⇒ AB = √ ((3)2 + (4)2)


⇒ AB = √ (9 + 16)


⇒ AB = √ 25


⇒ AB = 5


Distance of BC


⇒ BC= √ ((0 – 3)2 + (8 – 4)2)


⇒ BC = √ ((–3)2 + (4)2)


⇒ BC = √ (9 + 16)


⇒ BC = √ 25


⇒ BC = 5


Distance of CD


⇒ CD = √ ((–3 – 0)2 + (4 – 8)2)


⇒ CD = √ ((–3)2 + (–4)2)


⇒ CD = √ (9 + 16)


⇒ CD = √25


⇒ CD = 5


Distance of AD


⇒ AD = √ ((–3 – 0)2 + (4 – 0)2)


⇒ AD = √ ((–3)2 + (4)2)


⇒ AD = √ (9 + 16)


⇒ AD = √ 25


⇒ AD = 5


Distance of AC


⇒ AC = √ ((0 – 0)2 + (8 – 0)2)


⇒ AC = √ ((0)2 + (8)2)


⇒ AC = √ (64)


⇒ AC = 8


Distance of BD


⇒ BD = √ ((–3 – 3)2 + (4 – 4)2)


⇒ BD = √ ((–6)2 + (0)2)


⇒ BD = √ (36 +0)


⇒ BD = √ 36


⇒ BD = 6


AB = BC = CD = DA = 5 (That is, all the sides are equal.)


AC ≠ BD (That is, the diagonals are not equal.)


Hence the points A, B, C and D form a rhombus.



Question 37.

Show that the following points taken in order form the vertices of a rhombus.

(−4, −7), (−1, 2), (8, 5) and (5, −4)


Answer:

Formula used: 


(–4, –7), (–1, 2), (8, 5) and (5, –4)


Let the vertices be taken as A (–4,–7), B (–1, 2), C (8, 5) and D (5, –4).


Distance of AB


⇒ AB = √ ((–1 – (–4))2 + (2 – (–7)2))


⇒ AB = √ ((–1+4)2 + (2+7)2)


⇒ AB = √ ((3)2 + (9)2)


⇒ AB = √ (9 + 81)


⇒ AB = √ 100


⇒ AB = 10


Distance of BC


⇒ BC= √ ((8 – (–1))2 + (5 – 2)2)


⇒ BC = √ ((8+1)2 + (3)2)


⇒ BC = √ ((9)2+ 9)


⇒ BC = √ (81 + 9)


⇒ BC = √ 100


⇒ BC = 10


Distance of CD


⇒ CD = √ ((5 – 8)2 + (–4 –5 )2)


⇒ CD = √ ((3)2 + (–9)2)


⇒ CD = √ (9 + 81)


⇒ CD = √100


⇒ CD = 10


Distance of AD


⇒ AD = √ ((5 – (–4))2 + (–4 –(–7) )2)


⇒ AD = √ ((5+4)2 + (–4+7)2)


⇒ AD = √ ((9)2 +(3)2)


⇒ AD = √ (81+9)


⇒ AD = √ 100


⇒ AD = 10


Distance of AC


⇒ AC = √ ((8 – (–4))2 + (5 – (–7))2)


⇒ AC = √ ((8+4)2 + (5+7)2)


⇒ AC = √ ((12)2 +(12)2)


⇒ AC = √ (144 + 144)


⇒ AC = √ (288)


Distance of BD


⇒ BD = √ ((5 – (–1))2 + (–4 – 2)2)


⇒ BD = √ ((5 + 1))2 + (–4 – 2)2)


⇒ BD = √ ((6)2 + (–6)2)


⇒ BD = √ (36 + 36)


⇒ BD = √ 72


AB = BC = CD = DA = 10 (That is, all the sides are equal.)


AC ≠ BD (That is, the diagonals are not equal.)


Hence the points A, B, C and D form a rhombus.



Question 38.

Show that the following points taken in order form the vertices of a rhombus.

(1, 0), (5, 3), (2, 7) and (−2, 4)


Answer:

Formula used: 


(1, 0), (5, 3), (2, 7) and (–2, 4)


Let the vertices be taken as A (1, 0), B (5, 3), C (2, 7) and D (–2, 4).


Distance of AB


⇒ AB = √ ((5 – 1)2 + (3 – 0)2)


⇒ AB = √ ((4)2 + (3)2)


⇒ AB = √ (16 + 9)


⇒ AB = √ 25


⇒ AB = 5


Distance of BC


⇒ BC= √ ((2 – 5)2 + (7 – 3)2)


⇒ BC = √ ((3)2 + (4)2)


⇒ BC = √ (9 + 16)


⇒ BC = √ 25


⇒ BC = 5


Distance of CD


⇒ CD = √ ((–2 – 2)2 + (4 – 7)2)


⇒ CD = √ ((–4)2 + (–3)2)


⇒ CD = √ (16 + 9)


⇒ CD = √25


⇒ CD = 5


Distance of AD


⇒ AD = √ ((–2 – 1)2 + (4 – 0)2)


⇒ AD = √ ((–3)2 +(4)2)


⇒ AD = √ (9 + 16)


⇒ AD = √ 25


⇒ AD = 5


Distance of AC


⇒ AC = √ ((2 – 1)2 + (7 – 0)2)


⇒ AC = √ ((1)2 + (7)2)


⇒ AC = √ (1 + 49)


⇒ AC = √ 50


Distance of BD


⇒ BD = √ ((–2 – 5)2 + (4 – 3)2)


⇒ BD = √ ((–7)2 + (1)2)


⇒ BD = √ (49 + 1)


⇒ BD = √ 50


AB = BC = CD = DA = 10 (That is, all the sides are equal.)


AC ≠ BD (That is, the diagonals are not equal.)


Hence the points A, B, C and D form a rhombus.



Question 39.

Show that the following points taken in order form the vertices of a rhombus.

(2, −3), (6, 5), (−2, 1) and (−6, −7)


Answer:

Formula used: 


(2, –3), (6, 5), (–2, 1) and (–6, –7)


Let the vertices be taken as A (2, –3), B (6, 5), C (–2, 1) and D (–6, –7).


Distance of AB


⇒ AB = √ ((6 – 2)2 + (5 – (–3)2))


⇒ AB = √ ((6 – 2)2 + (5 + 3)2)


⇒ AB = √ ((4)2 + (8)2)


⇒ AB = √ (16 + 64)


⇒ AB = √ 80


Distance of BC


⇒ BC= √ ((–2 – 6)2 + (1 – 5)2)


⇒ BC = √ ((–8)2 + (–4)2)


⇒ BC = √ (64 + 16)


⇒ BC = √ 80


Distance of CD


⇒ CD = √ ((–6 – (–2))2 + (–7 – 1)2)


⇒ CD = √ ((–6 + 2)2 + (–7 – 1)2)


⇒ CD = √ ((–4)2 + (–8)2)


⇒ CD = √ (16 + 64)


⇒ CD = √80


Distance of AD


⇒ AD = √ ((–6 – (2))2 + (–7 – (–3))2)


⇒ AD = √ ((–6 – 2)2 + (–7 + 3)2)


⇒ AD = √ ((–8)2 +(–4)2)


⇒ AD = √ (64 + 16)


⇒ AD = √ 80


Distance of AC


⇒ AC = √ ((–2 – 2)2 + (1 – (–3))2)


⇒ AC = √ ((–2 – 2)2 + (1 + 3)2)


⇒ AC = √ ((–4)2 +(4)2)


⇒ AC = √ (16 + 16)


⇒ AC = √ 32


Distance of BD


⇒ BD = √ ((–6 – 6)2 + (–7 – 5)2)


⇒ BD = √ ((–6 – 6))2 + (–7 – 5)2)


⇒ BD = √ ((–12)2 + (–12)2)


⇒ BD = √ (144 + 144)


⇒ BD = √ 288


AB = BC = CD = DA = √80 (That is, all the sides are equal.)


AC ≠ BD (That is, the diagonals are not equal.)


Hence the points A, B, C and D form a rhombus.



Question 40.

Show that the following points taken in order form the vertices of a rhombus.

(15, 20), (−3, 12), (−11, −6) and (7, 2)


Answer:

Formula used: 


(15, 20), (–3, 12), (–11, –6) and (7, 2)


Let the vertices be taken as A (15, 20), B (–3, 12), C (–11, –6) and D (7, 2).


Distance of AB


⇒ AB = √ ((–3 – 15)2 + (12 – 20)2)


⇒ AB = √ ((–18)2 + (–8)2)


⇒ AB = √ (324 + 64)


⇒ AB = √ 388


Distance of BC


⇒ BC= √ ((–11 –(–3))2 + (–6 – 12)2)


⇒ BC = √ (–11 + 3)2 + (–6 – 12)2)


⇒ BC = √ ((–8)2 + (–18)2)


⇒ BC = √ (64 + 324)


⇒ BC = √ 388


Distance of CD


⇒ CD = √ ((7 – (–11))2 + (2 – (–6))2)


⇒ CD = √ ((7 + 11)2 + (2 + 6)2)


⇒ CD = √ ((18)2 + (8)2)


⇒ CD = √ (324 + 64)


⇒ CD = √388


Distance of AD


⇒ AD = √ ((7 – 15))2 + (2 – 20)2)


⇒ AD = √ ((–8)2 +(–18)2)


⇒ AD = √ (64 + 324)


⇒ AD = √ 388


Distance of AC


⇒ AC = √ ((–11 – 15)2 + (–6 – 20)2)


⇒ AC = √ ((–26)2 +(–26)2)


⇒ AC = √ (676 + 676)


⇒ AC = √ 1352


Distance of BD


⇒ BD = √ ((7 – (–3))2 + (2 – 12)2)


⇒ BD = √ ((7 + 3))2 + (2 – 12)2)


⇒ BD = √ ((10)2 + (–10)2)


⇒ BD = √ (100 + 100)


⇒ BD = √ 200


AB = BC = CD = DA = √388 (That is, all the sides are equal.)


AC ≠ BD (That is, the diagonals are not equal.)


Hence the points A, B, C and D form a rhombus.



Question 41.

Examine whether the following points taken in order form a square.

(0, −1), (2, 1), (0, 3) and (−2, 1)


Answer:

Formula used: 


(0, –1), (2, 1), (0, 3) and (–2, 1)


Let the vertices be taken as A (0, –1), B (2, 1), C (0, 3) and D (–2, 1).


Distance of AB


⇒ AB = √ ((2 – 0)2 + ((1 – (–1))2)


⇒ AB = √ ((2 – 0))2 + (1 + 1)2)


⇒ AB = √ ((2)2 + (2)2)


⇒ AB = √ (4 + 4)


⇒ AB = √ 8


Distance of BC


⇒ BC= √ ((0 – 2)2 + (3 – 1)2)


⇒ BC = √ ((–2)2 + (2)2)


⇒ BC = √ (4 + 4)


⇒ BC = √ 8


Distance of CD


⇒ CD = √ ((–2 – 0)2 + (1 – 3)2)


⇒ CD = √ ((–2)2 + (–2)2)


⇒ CD = √ (4 + 4)


⇒ CD = √8


Distance of AD


⇒ AD = √ ((–2 – 0)2 + (1 – (–1))2)


⇒ AD = √ ((–2 – 0)2 + (1 + 1)2)


⇒ AD = √ ((–2)2 + (2)2)


⇒ AD = √ (4 + 4)


⇒ AD = √ 8


Distance of AC


⇒ AC = √ ((0 – 0)2 + (3 – (–1))2)


⇒ AC = √ ((0 – 0)2 + (3 + 1)2)


⇒ AC = √ ((0)2 + (4)2)


⇒ AC = √ (0 + 16)


⇒ AC = √ 16


⇒ AC = 4


Distance of BD


⇒ AC = √ ((–2 – 2)2 + (1 – 1)2)


⇒ AC = √ ((–4)2 + (0)2)


⇒ AC = √ (16 + 0)


⇒ AC = √ 16


⇒ AC = 4


AB = BC = CD = DA = √8 (That is, all the sides are equal.)


AC = BD = 4. (That is, the diagonals are equal.)


Hence the points A, B, C and D form a square.



Question 42.

Examine whether the following points taken in order form a square.

(5, 2), (1, 5), (−2, 1) and (2, −2)


Answer:

Formula used: 


(5, 2), (1, 5), (–2, 1) and (2, –2)


Let the vertices be taken as A (5, 2), B (1, 5), C (–2, 1) and D (2, –2).


Distance of AB


⇒ AB = √ ((1 – 5)2 + ((5 – 2)2)


⇒ AB = √ ((–4)2 + (3)2)


⇒ AB = √ (16 + 9)


⇒ AB = √25


⇒ AB = 5


Distance of BC


⇒ BC= √ ((–2 – 1)2 + (1 – 5)2)


⇒ BC = √ ((–3)2 + (–4)2)


⇒ BC = √ (9 + 16)


⇒ BC = √ 25


⇒ BC = 5


Distance of CD


⇒ CD = √ ((2 – (–2))2 + (–2 – 1)2)


⇒ CD = √ ((2 + 2)2 + (–2 – 1)2)


⇒ CD = √ ((4)2 + (–3)2)


⇒ CD = √ (16 + 9)


⇒ CD = √25


⇒ CD = 5


Distance of AD


⇒ AD = √ ((2 – 5)2 + (–2 – 2)2)


⇒ AD = √ ((–3)2 + (–4)2)


⇒ AD = √ (9 + 16)


⇒ AD = √ 25


⇒ AD = 5


Distance of AC


⇒ AC = √ ((–2 – 5)2 + (1 – 2)2)


⇒ AC = √ ((–7)2 + (–1)2)


⇒ AC = √ (49 + 1)


⇒ AC = √50


⇒ AC = 5√2


Distance of BD


⇒ BD = √ ((2 – 1)2 + (–2 – 5)2)


⇒ BD = √ ((1)2 + (–7)2)


⇒ BD = √ (1 + 49)


⇒ BD = √ 50


⇒ BD = 5√2


AB = BC = CD = DA = 5 (That is, all the sides are equal.)


AC = BD = 5√2. (That is, the diagonals are equal.)


Hence the points A, B, C and D form a square.



Question 43.

Examine whether the following points taken in order form a square.

(3, 2), (0, 5), (−3, 2) and (0, −1)


Answer:

Formula used: 


(3, 2), (0, 5), (–3, 2) and (0, –1)


Let the vertices be taken as A (3, 2), B (0, 5), C (–3, 2) and D (0, –1).


Distance of AB


⇒ AB = √ ((0 – 3)2 + ((5 – 2)2)


⇒ AB = √ ((–3)2 + (3)2)


⇒ AB = √ (9 + 9)


⇒ AB = √18


Distance of BC


⇒ BC= √ ((–3 – 0)2 + (2 – 5)2)


⇒ BC = √ ((–3)2 + (–3)2)


⇒ BC = √ (9 + 9)


⇒ BC = √ 18


Distance of CD


⇒ CD = √ ((0 – (–3))2 + (–1 – 2)2)


⇒ CD = √ ((0 + 3)2 + (–1 – 2)2)


⇒ CD = √ ((3)2 + (–3)2)


⇒ CD = √ (9 + 9)


⇒ CD = √18


Distance of AD


⇒ AD = √ ((0 – 3)2 + (–1 – 2)2)


⇒ AD = √ ((–3)2 + (–3)2)


⇒ AD = √ (9 + 9)


⇒ AD = √ 18


Distance of AC


⇒ AC = √ ((–3 – 3)2 + (2 – 2)2)


⇒ AC = √ ((–6)2 + (0)2)


⇒ AC = √ (36 + 0)


⇒ AC = √36


⇒ AC = 6


Distance of BD


⇒ BD = √ ((0 – 0)2 + (–1 – 5)2)


⇒ BD = √ ((0)2 + (–6)2)


⇒ BD = √ (0 + 36)


⇒ BD = √ 36


⇒ BD = 6


AB = BC = CD = DA = √18. (That is, all the sides are equal.)


AC = BD = 6. (That is, the diagonals are equal.)


Hence the points A, B, C and D form a square.



Question 44.

Examine whether the following points taken in order form a square.

(12, 9), (20, −6), (5, −14) and (−3, 1)


Answer:

Formula used: 


(12, 9), (20, –6), (5, –14) and (–3, 1)


Let the vertices be taken as A (12, 9), B (20, –6), C (5, –14) and D (–3, 1).


Distance of AB


⇒ AB = √ ((20 – 12)2 + ((–6 – 9)2)


⇒ AB = √ ((8)2 + (–15)2)


⇒ AB = √ (64 + 225)


⇒ AB = √289


Distance of BC


⇒ BC= √ ((5 – 20)2 + (–14 – (–6))2)


⇒ BC= √ ((5 – 20)2 + (–14 + 6)2)


⇒ BC = √ ((–15)2 + (–8)2)


⇒ BC = √ (225 + 64)


⇒ BC = √ 289


Distance of CD


⇒ CD = √ ((–3 – 5)2 + (1 – (–14))2)


⇒ CD = √ ((–3 – 5)2 + (1 + 14)2)


⇒ CD = √ ((–8)2 + (15)2)


⇒ CD = √ (64 + 225)


⇒ CD = √289


Distance of AD


⇒ AD = √ ((–3 – 12)2 + (1 – 9)2)


⇒ AD = √ ((–15)2 + (–8)2)


⇒ AD = √ (225 + 64)


⇒ AD = √ 289


Distance of AC


⇒ AC = √ ((5 – 12)2 + (–14 – 9)2)


⇒ AC = √ ((–7)2 + (–23)2)


⇒ AC = √ (49 + 529)


⇒ AC = √578


Distance of BD


⇒ BD = √ ((–3 – 20)2 + (1 – (–6))2)


⇒ BD = √ ((–3 – 20)2 + (1 + 6)2)


⇒ BD = √ ((–23)2 + (7)2)


⇒ BD = √ (529 + 49)


⇒ BD = √ 578


AB = BC = CD = DA = √ 289 (That is, all the sides are equal.)


AC = BD = √578. (That is, the diagonals are equal.)


Hence the points A, B, C and D form a square.



Question 45.

Examine whether the following points taken in order form a square.

(−1, 2), (1, 0), (3, 2) and (1, 4)


Answer:

Formula used: 


(–1, 2), (1, 0), (3, 2) and (1, 4)


Let the vertices be taken as A (–1, 2), B (1, 0), C (3, 2) and D (1, 4).


Distance of AB


⇒ AB = √ ((1 – (–1))2 + ((0 – 2)2)


⇒ AB = √ ((1 + 1)2 + (0 – 2)2)


⇒ AB = √ ((2)2 + (–2)2)


⇒ AB = √ (4 + 4)


⇒ AB = √8


Distance of BC


⇒ BC= √ ((3 – 1)2 + (2 – 0)2)


⇒ BC = √ ((2)2 + (2)2)


⇒ BC = √ (4 + 4)


⇒ BC = √ 8


Distance of CD


⇒ CD = √ ((1 – 3)2 + (4 – 2))2)


⇒ CD = √ ((–2)2 + (2)2)


⇒ CD = √ (4 + 4)


⇒ CD = √8


Distance of AD


⇒ AD = √ ((1 – (–1))2 + (4 – 2)2)


⇒ AD = √ ((1 + 1)2 + (4 – 2)2)


⇒ AD = √ ((2)2 + (2)2)


⇒ AD = √ (4 + 4)


⇒ AD = √ 8


Distance of AC


⇒ AC = √ ((3 – (–1))2 + (2 – 2)2)


⇒ AC = √ ((3 + 1) + (2 – 2)2)


⇒ AC = √ ((4)2 + (0)2)


⇒ AC = √ (16 + 0)


⇒ AC = √16


⇒ AC = 4


Distance of BD


⇒ BD = √ ((1 – 1)2 + (4 – 0)2)


⇒ BD = √ ((0)2 + (4)2)


⇒ BD = √ (0 + 16)


⇒ BD = √16


⇒ BD = 4


AB = BC = CD = DA = √8 (That is, all the sides are equal.)


AC = BD = 4. (That is, the diagonals are equal.)


Hence the points A, B, C and D form a square.



Question 46.

Examine whether the following points taken in order form a rectangle.

(8, 3), (0, −1), (−2, 3) and (6, 7)


Answer:

Formula used: 


(8, 3), (0, –1), (–2, 3) and (6, 7)


Let the vertices be taken as A (8, 3), B (0, –1), C (–2, 3) and D (6, 7).


Distance of AB


⇒ AB = √ ((0 – 8)2 + ((–1 – 3)2)


⇒ AB = √ ((–8)2 + (–4)2)


⇒ AB = √ (64 + 16)


⇒ AB = √ 80


Distance of BC


⇒ BC= √ ((–2 – 0)2 + (3 – (–1))2)


⇒ BC = √ ((–2 – 0)2 + (3 + 1)2)


⇒ BC = √ ((–2)2 + (4)2)


⇒ BC = √ (4 + 16)


⇒ BC = √ 20


Distance of CD


⇒ CD = √ ((6 – (–2))2 + (7 – 3)2)


⇒ CD = √ ((6 + 2)2 + (7 – 3)2)


⇒ CD = √ ((8)2 + (4)2)


⇒ CD = √ (64 + 16)


⇒ CD = √80


Distance of AD


⇒ AD = √ ((6 – 8)2 + (7 – 3)2)


⇒ AD = √ ((–2)2 + (4)2)


⇒ AD = √ (4 + 16)


⇒ AD = √ 20


Distance of AC


⇒ AC = √ ((–2 – 8)2 + (3 – 3)2)


⇒ AC = √ ((–10)2 + (0)2)


⇒ AC = √ (100 + 0)


⇒ AC = √ 100


⇒ AC = 10


Distance of BD


⇒ BD = √ ((6 – 0 )2 + (7 – (–1))2)


⇒ BD = √ ((6 – 0)2 + (7 + 1)2)


⇒ BD = √ ((6)2 + (8)2)


⇒ BD = √ (36 + 64)


⇒ BD = √ 100


⇒ BD = 10


AB = CD = √80 and BC = AD = √ 20 (opposite sides of rectangle are equal).


AC = BD = 10 (Diagonals of rectangle are equal)


Hence the points A, B, C and D form a square.



Question 47.

Examine whether the following points taken in order form a rectangle.

(−1, 1), (0, 0), (3, 3) and (2, 4)


Answer:

Formula used: 


(–1, 1), (0, 0), (3, 3) and (2, 4)


Let the vertices be taken as A (–1, 1), B (0, 0), C (3, 3) and D (2, 4).


Distance of AB


⇒ AB = √ ((0 – (–1))2 + (0 – 1)2)


⇒ AB = √ ((0 + 1)2 + (0 – 1)2)


⇒ AB = √ ((1)2 + (–1)2)


⇒ AB = √ (1 + 1)


⇒ AB = √ 2


Distance of BC


⇒ BC= √ ((3 – 0)2 + (3 – 0)2)


⇒ BC = √ ((3)2 + (3)2)


⇒ BC = √ (9 + 9)


⇒ BC = √ 18


Distance of CD


⇒ CD = √ ((2 – 3)2 + (4 – 3)2)


⇒ CD = √ ((1)2 + (1)2)


⇒ CD = √ (1 + 1)


⇒ CD = √2


Distance of AD


⇒ AD = √ ((2 – (–1))2 + (4 – 1)2)


⇒ AD = √ ((2 + 1)2 + (4 – 1)2)


⇒ AD = √ ((3)2 + (3)2)


⇒ AD = √ (9 + 9)


⇒ AD = √ 18


Distance of AC


⇒ AC = √ ((3 – (–1))2 + (3 – 1)2)


⇒ AC = √ ((3 + 1)2 + (3 – 1)2)


⇒ AC = √ ((4)2 + (2)2)


⇒ AC = √ (16 + 4)


⇒ AC = √ 20


Distance of BD


⇒ BD = √ ((2 – 0)2 + (4 – 0)2)


⇒ BD = √ ((2)2 + (4)2)


⇒ BD = √ (4 + 16)


⇒ BD = √ 20


AB = CD = √2 and BC = AD = √ 18 (opposite sides of rectangle are equal).


AC = BD = √ 20 (Diagonals of rectangle are equal)


Hence the points A, B, C and D form a square.



Question 48.

Examine whether the following points taken in order form a rectangle.

(−3, 0), (1, −2), (5, 6) and (1, 8)


Answer:

Formula used: 


(–3, 0), (1, –2), (5, 6) and (1, 8)


Let the vertices be taken as A (–3, 0), B (1, –2), C (5, 6) and D (1, 8).


Distance of AB


⇒ AB = √ ((1 – (–3))2 + ((–2 – 0)2)


⇒ AB = √ ((1 + 3)2 + (–2 – 0)2)


⇒ AB = √ ((4)2 + (–2)2)


⇒ AB = √ (16 + 4)


⇒ AB = √ 20


Distance of BC


⇒ BC= √ ((5 – 1)2 + (6 – (–2))2)


⇒ BC = √ ((5 – 1)2 + (6 + 2)2)


⇒ BC = √ ((4)2 + (8)2)


⇒ BC = √ (16 + 64)


⇒ BC = √ 80


Distance of CD


⇒ CD = √ ((1 – 5)2 + (8 – 6)2)


⇒ CD = √ ((–4)2 + (2)2)


⇒ CD = √ (16 + 4)


⇒ CD = √ 20


Distance of AD


⇒ AD = √ ((1 – (–3))2 + (8 – 0)2)


⇒ AD = √ ((1 + 3)2 + (8 – 0)2)


⇒ AD = √ ((4)2 + (8)2)


⇒ AD = √ (16 + 64)


⇒ AD = √ 80


Distance of AC


⇒ AC = √ ((5 – (–3))2 + (6 – 0)2)


⇒ AC = √ ((5 + 3)2 + (6 – 0)2)


⇒ AC = √ ((8)2 + (6)2)


⇒ AC = √ (64 + 36)


⇒ AC = √ 100


⇒ AC = 10


Distance of BD


⇒ BD = √ ((1 – 1)2 + (8 – (–2))2)


⇒ BD = √ ((1 – 1)2 + (8 + 2)2)


⇒ BD = √ ((0)2 + (10)2)


⇒ BD = √ (0 + 100)


⇒ BD = √ 100


⇒ BD = 10


AB = CD = √20 and BC = AD = √ 80 (opposite sides of rectangle are equal).


AC = BD = 10 (Diagonals of rectangle are equal)


Hence the points A, B, C and D form a square.



Question 49.

If the distance between two points (x,7) and (1, 15) is 10, find x.


Answer:

Formula used: 


Given: Distance = 10 and coordinates of two points is A (x, 7) and B (1, 15)


AB = √ (x2 – x1)2 + (y2 – y1)2


⇒ 10 = √ (1 – x)2 + (15 – 7)2


⇒ 10 = √ (1 – x)2 + 82


Squaring both sides


⇒ 102 = (1 – x)2 + 82


⇒ 100 = 1 – 2x + x2 + 64


⇒ 100 = x2 – 2x + 65


⇒ x2 – 2x + 65 – 100 = 0


⇒ x2 – 2x – 35 = 0


⇒ x2 – 7x + 5x – 35 = 0


⇒ x (x – 7) + 5(x – 7) = 0


⇒ (x – 7) (x + 5) = 0


x – 7 = 0 or x + 5 = 0


x = 7 or x = –5



Question 50.

Show that (4, 1) is equidistant from the points (−10, 6) and (9, −13).


Answer:

Formula used:


Let the points be A (4, 1), B (–10, 6) and C (9, –13)


Distance of AB


⇒ AB = √ ((–10 – 4)2 + (6 – 1)2)


⇒ AB = √ ((–14)2 + (5)2)


⇒ AB = √ (196 + 25


⇒ AB = √ 221


Distance of BC


⇒ BC = √ ((9 – 4)2 + (–13 – 1)2)


⇒ BC = √ ((5)2 + (–14)2)


⇒ BC = √ (25 + 196


⇒ BC = √ 221


∴ AB = BC = √ 221



Question 51.

If two points (2, 3) and (−6, −5) are equidistant from the point (x, y), show that x + y + 3 = 0.


Answer:

Formula used: 


Let the points be A (x, y), B (2, 3) and C (–6, –5)


Distance of AB


⇒ AB = √ ((2 – x)2 + (3 – y)2)


⇒ AB = √ ((4 – 4x + x2) + (9 – 6y + y2))


⇒ AB = √ (4 – 4x + x2 + 9 – 6y + y2)


⇒ AB = √ x2 + y2 – 4x – 6y + 13


Distance of BC


⇒ BC = √ ((–6 – x)2 + (–5 – y)2)


⇒ BC = √ ((36 + x2 + 12x) + (25 + y2 + 10y))


⇒ BC = √ (36 + x2 + 12x + 25 + y2 + 10y)


⇒ BC = √ (x2 + y2 + 12x + 10y + 61)


i.e. AB = BC (∵ Given)


⇒ √x2 + y2 – 4x – 6y + 13 = √ x2 + y2 + 12x + 10y + 61


Squaring both sides


⇒ x2 + y2 – 4x – 6y + 13 = x2 + y2 + 12x + 10y + 61


⇒ x2 + y2 – 4x – 6y + 13 – x2 – y2 – 12x – 10y – 61 = 0


⇒ –16x – 16 y – 48 = 0


⇒ –4(x + y + 3) = 0


⇒ x + y + 3 = 0


Hence proved.



Question 52.

If the length of the line segment with end points (2, −6) and (2, y) is 4, find y.


Answer:

Formula used: 


Given: Distance = 4 and coordinates of two points is A (2, –6) and B (2, y)


AB = √ (x2 – x1)2 + (y2 – y1)2


⇒ 4 = √ (2 – 2)2 + (y – (–6))2


⇒ 4 = √ (0) + (y + 6)2


Squaring both sides


⇒ 42 = (y + 6)2


⇒ 16 = y2 + 12y + 36


⇒ y2 + 12y + 36 – 16 = 0


⇒ y2 + 12y + 20 = 0


⇒ y2 + 10y + 2y + 20 = 0


⇒ y (y + 10) + 2(y + 10) = 0


⇒ (y + 2) (y + 10) = 0


y + 2 = 0 or y + 10 = 0


y = –2 or y = –10


∴ y = –2 or –10



Question 53.

Find the perimeter of the triangle with vertices (i) (0, 8), (6, 0) and origin; (ii) (9, 3), (1, −3) and origin.


Answer:

Formula used: 


i). (0, 8), (6, 0) and (0, 0)


Let the points be A (0, 8), B (6, 0) and C (0, 0)


Distance of AB


⇒ AB = √ ((6 – 0)2 + (0 – 8)2)


⇒ AB = √ ((6)2 + (–8)2)


⇒ AB = √ (36 + 64)


⇒ AB = √ 100


⇒ AB = 10


Distance of BC


⇒ BC = √ ((0 – 6)2 + (0 – 0)2)


⇒ BC = √ ((–6)2 + (0)2)


⇒ BC = √ (36 + 0)


⇒ BC = √ 36


⇒ BC = 6


Distance of AC


⇒ AC = √ ((0 – 0)2 + (0 – 8)2)


⇒ AC = √ ((0)2 + (–8)2)


⇒ AC = √ (0 + 64)


⇒ AC = √ 64


⇒ AC = 8


Perimeter of ΔABC = AB + BC + AC


= 10 + 6 + 8


= 24


ii). (9, 3), (1, –3) and (0, 0)


Let the points be A (9, 3), B (1, –3) and C (0, 0)


Distance of AB


⇒ AB = √ ((1 – 9)2 + (–3 – 3)2)


⇒ AB = √ ((–8)2 + (–6)2)


⇒ AB = √ (64 + 36)


⇒ AB = √ 100


⇒ AB = 10


Distance of BC


⇒ BC = √ ((0 – 1)2 + (0 – (–3))2)


⇒ BC = √ ((0 – 1)2 + (0 + 3)2)


⇒ BC = √ ((–1)2 + (3)2)


⇒ BC = √ (1 + 9)


⇒ BC = √10


Distance of AC


⇒ AC = √ ((0 – 9)2 + (0 – 3)2)


⇒ AC = √ ((–9)2 + (–3)2)


⇒ AC = √ (81 + 8)


⇒ AC = √ 90


⇒ AC = 3√10


Perimeter of ΔABC = AB + BC + AC


= 10 + √ 10 + 3√ 10


= 10 + 4√10



Question 54.

Find the point on the y–axis equidistant from (−5, 2) and (9, −2) (Hint: A point on the y–axis will have its x–coordinate as zero).


Answer:

Formula used: 


Let the point A (–5, 2), B (9, –2) and C be the point on y–axis i.e. (0, y)


Distance of AC


⇒ AC = √ ((0 – (–5))2 + (y – 2)2)


⇒ AC = √ ((0 + 5)2 + (y – 2)2)


⇒ AC = √ ((5)2 + (y – 2)2)


⇒ AC = √ (25 + y2 – 4y + 4)


⇒ AC = √ y2 – 4y + 29


Distance of BC


⇒ BC = √ ((0 – 9)2 + (y – (–2)2)


⇒ BC = √ ((0 – 9)2 + (y + 2)2)


⇒ BC = √ ((9)2 + (y + 2)2)


⇒ BC = √ (81 + y2 + 4y + 4)


⇒ BC = √ y2 + 4y + 85


i.e. AC = BC (∵ Given)


⇒ √ y2 – 4y + 29 = √ y2 + 4y + 85


Squaring both sides


⇒ y2 – 4y + 29 = y2 + 4y + 8


⇒ y2 – 4y + 29 – y2 – 4y – 85 = 0


⇒ –8y – 56 = 0


⇒ –8 (y + 7) = 0


⇒ y + 7 = 0


y = –7


∴ the point on y–axis is (0, –7).



Question 55.

Find the radius of the circle whose center is (3, 2) and passes through (−5, 6).


Answer:

Formula used: 


Let the point be A (–5, 6) and O (3, 2)


Distance of OA


⇒ OA = √ ((–5 – 3)2 + (6 – 2)2)


⇒ OA = √ ((–8)2 + (4)2)


⇒ OA = √ (64 + 16)


⇒ OA = √ 80


⇒ OA = 4√5



Question 56.

Prove that the points (0, −5), (4, 3) and (−4, −3) lie on the circle centered at the origin with radius 5.


Answer:

Formula used: 


Let the point A (0, –5), B (4, 3) and C (–4, –3) lie on the circle with center O (0, 0)


Distance of AO


⇒ AO = √ ((0 – 0)2 + (0 – (–5))2)


⇒ AO = √ ((0 – 0)2 + (0 + 5)2)


⇒ AO = √ ((0)2 + (5)2)


⇒ AO = √ (0 + 25)


⇒ AO = √ 25


⇒ AO = 5


Distance of BO


⇒ BO = √ ((0 – 4)2 + (0 – 3)2)


⇒ BO = √ ((–4)2 + (–3)2)


⇒ BO = √ (16 + 9)


⇒ BO = √ 25


⇒ BO = 5


Distance of CO


⇒ CO = √ ((0 – (–4))2 + (0 – (–3))2)


⇒ CO = √ ((0 + 4)2 + (0 + 3)2)


⇒ CO = √ ((4)2 + (3)2)


⇒ CO = √ (16 + 9)


⇒ CO = √ 25


⇒ CO = 5


∴ AO = BO = CO = 5 = Radius


Hence, point A, B and C lie on the circle.



Question 57.

In the Fig. 5.20, PB is perpendicular segment from the point A (4, 3). If PA = PB then find the coordinates of B.



Answer:

Formula used: 


Let the point P (4, 0)


PB is perpendicular segment from point A to B


∴ let B be (4, –y)


Distance of PA


⇒ PA = √ ((4 – 4)2 + (3 – 0)2)


⇒ PA = √ ((0)2 + (3)2)


⇒ PA = √ (0 + 9)


⇒ PA= √ 9


⇒ PA = 3


Distance of PB


⇒ PB = √ ((4 – 4)2 + (–y – 0)2)


⇒ PB = √ ((4 – 4)2 + (–y)2)


⇒ PB = √ ((0)2 + (–y)2)


⇒ PB = √ 0 + y2


⇒ PB = y2


i.e. AP = BP


⇒ 3 = √ y2


Squaring both sides


⇒ 9 = y2


⇒ y = √9


⇒ y = 3


∴ Point B is (4, –3)



Question 58.

Find the area of the rhombus ABCD with vertices A (2, 0), B (5, –5), C (8, 0) and D (5, 5). [Hint: Area of the rhombus ABCD = 1/2d1 d2]


Answer:

Formula used: 


Coordinates of rhombus are A (2, 0), B (5, –5), C (8, 0) and D (5, 5)


Area of rhombus = 


Distance of AC(d1)


⇒ AC = √ ((8 – 2)2 + (0 – 0)2)


⇒ AC = √ ((6)2 + (0)2)


⇒ AC = √ (36 + 0)


⇒ AC = √ 36


⇒ AC = 6


Distance of BD(d2)


⇒ BD = √ ((5 – 5)2 + (5 – (–5))2)


⇒ BD = √ ((5 – 5)2 + (5 + 5)2)


⇒ BD = √ ((0)2 + (10)2)


⇒ BD = √ (0 + 100)


⇒ BD = √ 100


⇒ BD = 10


∴ Area of rhombus = 


⇒ Area 


⇒ Area = 3 × 10


⇒ Area = 30 units sq.



Question 59.

Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14)? Give reason.


Answer:

Formula used: 


Let the points A (1, 5) B (5, 8) and C (13, 14)


Distance of AB


⇒ AB = √ ((5 – 1)2 + (8 – 5)2)


⇒ AB = √ ((4)2 + (3)2)


⇒ AB = √ (16 + 9)


⇒ AB = √ 25


⇒ AB = 5


Distance of BC


⇒ BC = √ ((13 – 5)2 + (14 – 8)2)


⇒ BC = √ ((8)2 + (6)2)


⇒ BC = √ (64 + 36)


⇒ BC = √ 100


⇒ BC = 10


Distance of AC


⇒ AC = √ ((13 – 1)2 + (14 – 5)2)


⇒ AC = √ ((12)2 + (9)2)


⇒ AC = √ (144 + 81)


⇒ AC = √ 225


⇒ AC = 15


Now, we can see that AB + BC = AC.


∴ A, B and C are collinear. Hence, we cannot draw triangle using these coordinates.



Question 60.

If origin is the center of a circle with radius 17 units, find the coordinates of any four points on the circle which are not on the axes. (Use the Pythagorean triplets)


Answer:

Formula used: 


Let the point be A (x, y)


Center is at origin (0, 0)


Distance of OA


⇒ OA = √((x – 0)2 + (y – 0)2)


⇒ OA = √ ((x)2 + (y)2)


⇒ OA = √ x2 + y2


Squaring both sides


⇒ (0A)2 = x2 + y2


⇒ (17)2 = x2 + y2


Using Pythagorean triplet


x and y can 8 and 5 or vice–a–versa.


∴ x = ± 8 or ±15


y = ± 8 or ±15


Hence, coordinate on circle other than coordinates on axis are


(8, 15), (–8, –15), (–8, 15) and (8, –15)



Question 61.

Show that (2, 1) is the circum–center of the triangle formed by the vertices (3, 1), (2, 2) and (1, 1).


Answer:

Formula used: 


Let the points be A (3, 1), B (2, 2), C (1, 1) and S(2, 1)


Distance of SA


⇒ SA = √ ((3 – 2)2 + (1 – 1)2)


⇒ SA = √ ((1)2 + (0)2


⇒ SA = √ (1 + 0)


⇒ SA = √ 1 = 1


Distance of SB


⇒ SB = √ ((2 – 2)2 + (2 – 1)2)


⇒ SB = √ ((0)2 + (1)2


⇒ SB = √ (0 + 1)


⇒ SB = √ 1 = 1


Distance of SC


⇒ SC = √ ((1 – 2)2 + (1 – 1)2)


⇒ SC = √ ((–1)2 + (0)2


⇒ SC = √ (1 + 0)


⇒ SC = √ 1 = 1


It is known that the circum–centre is equidistant from all the vertices of a triangle.


Since S is equidistant from all the three vertices, it is the circum–centre of the triangle ABC.



Question 62.

Show that the origin is the circum–center of the triangle formed by the vertices (1, 0), (0, −1) and .


Answer:

Formula used: 


Let the points be A (1, 0), B (0, –1), C  and S (0, 0)


Distance of SA


⇒ SA = √ ((1 – 0)2 + (0 – 0)2)


⇒ SA = √ ((1)2 + (0)2


⇒ SA = √ (1 + 0)


⇒ SA = √ 1 = 1


Distance of SB


⇒ SB = √ ((0 – 0)2 + (–1 – 0)2)


⇒ SB = √ ((0)2 + (–1)2


⇒ SB = √ (0 + 1)


⇒ SB = √ 1 = 1


Distance of SC


⇒ SC 


⇒ SC 


⇒ SC 


⇒ SC 


⇒ SC =√ 1 = 1


It is known that the circum–centre is equidistant from all the vertices of a triangle.


Since S is equidistant from all the three vertices, it is the circum–centre of the triangle ABC.



Question 63.

If the points A (6, 1), B (8, 2), C (9, 4) and D (p, 3) taken in order are the vertices of a parallelogram, find the value of p using distance formula.


Answer:

Formula used: 


Let A, B, C and D represent the points (6, 1), (8, 2), (9, 4) and (p, 3)


Distance of AB


⇒ AB = √ ((8 – 6))2 + (2 – 1)2)


⇒ AB = √ ((2)2 + (1)2)


⇒ AB = √ (4 + 1)


⇒ AB = √ 5


Distance of CD


⇒ CD = √ ((p – 9)2 + (3 – 4)2)


⇒ CD = √ ((p – 9)2 + (1)2)


⇒ CD = √ (p2 + 81 – 18p + 1)


⇒ CD = √ p2 – 18p + 82


i.e., The opposite sides are equal.


∴ AB = CD


⇒ √5 = √ p2 – 18p + 82


Squaring both sides


⇒ 5 = p2 – 18p + 82


⇒ p2 – 18p + 82 – 5 =0


⇒ p2 – 18p + 77 = 0


⇒ p2 – 11p – 7p + 77 = 0


⇒ p(p – 11) – 7(p – 11)= 0


⇒ (p – 11)(p – 7) = 0


p – 11 = 0 or p – 7 = 0


p = 11 or p = 7



Question 64.

The radius of the circle with center at the origin is 10 units. Write the coordinates of the point where the circle intersects the axes. Find the distance between any two of such points.


Answer:

Formula used: 


Let the point be A (x, 0) and B (0, y)


Given center O (0, 0) and radius = 10


Distance of OA


⇒ 5 = √ ((x – 0)2 + (0 – 0)2)


⇒ 5 = √ ((x)2 + (0)2)


⇒ 5= √ (x2 + 0)


⇒ 5 = √ x2


⇒ 5 = x


∴ point A is (5, 0)


Distance of OB


⇒ 5 = √ ((0 – 0)2 + (y – 0)2)


⇒ 5 = √ ((0)2 + (y)2)


⇒ 5 = √ (0 + y2)


⇒ 5 = √ y2


⇒ 5 = y


∴ point B is (0, 5)


Now,


Distance AB = √ ((0 – 5)2 + (5 – 0)2)


= √ ((–5)2 + (5)2)


= √ (25 + 25)


= √ (50)


= 5√2




Exercise 5.3
Question 1.

The point (–2, 7) lies is the quadrant
A. I

B. II

C. III

D. IV


Answer:

Option A: value to lie in I quadrant, both should be positive. Hence, this is not correct.


Option B: value to lie in II quadrant, x–coordinate should be negative and y–coordinate should be positive. Hence, this is correct.


Option C: value to lie in III quadrant, both should be negative. Hence, this is not correct.


Option D: value to lie in I quadrant, x–coordinate should be positive and y– coordinate should be negative. Hence, this is not correct.


Question 2.

The point (x, 0) where x < 0 lies on
A. OX

B. OY

C. OX’

D. OY’


Answer:

Option A: point on OX, x–coordinate will be greater than 0 i.e. x > 0.


Option B: point on OY, y–coordinate will be greater than 0 i.e. y > 0.


Option C: point on OX’, x–coordinate will be lesser than 0 i.e. x < 0.


Option D: point on OY’, y–coordinate will be lesser than 0 i.e. y < 0.


Question 3.

For a point A (a, b) lying in quadrant III
A. a > 0, b < 0

B. a < 0, b < 0

C. a > 0, b > 0

D. a < 0, b > 0


Answer:

Option A: point with a > 0, b < 0, lies in the IV quadrant.


Option B: point with a < 0, b < 0, lies in the III quadrant.


Option C: point with a > 0, b > 0, lies in the I quadrant.


Option D: point with a < 0, b > 0, lies in the II quadrant.


Question 4.

The diagonal of a square formed by the points (1, 0) (0, 1) (–1, 0) and (0, –1) is
A. 2

B. 4

C. √2

D. 8


Answer:

Formula used: 


Let the points be A (1, 0), B (0, 1), C (–1, 0) and D (0, –1)


Distance of diagonal AC


⇒ AC = √ ((–1 – 1)2 + ( 0 – 0)2)


⇒ AC = √ ((–2)2 + (0)2)


⇒ AC =√ (4 + 0)


⇒ AC = √ 4


⇒ AC = 2


∴ Option A is correct.


Question 5.

The triangle obtained by joining the points A (–5, 0) B (5, 0) and C (0, 6) is
A. an isosceles triangle

B. right triangle

C. scalene triangle

D. an equilateral triangle


Answer:

Formula used: 


Let the point be A (–5, 0) B (5, 0) and C (0, 6)


Distance of AB


⇒ AB = √ (5 – (–5))2 + (0 – 0)2)


⇒ AB = √ (5 + 5)2 + (0 – 0)2)


⇒ AB = √ ((10)2 + (0)2)


⇒ AB = √ (100 + 0)


⇒ AB = √ 100 = 10


Distance of AC


⇒ AC = √ ((0 – (–5))2 + (6 – 0)2)


⇒ AC = √ ((5)2 + (6)2)


⇒ AC = √ (25 + 36)


⇒ AC = √ 61


Distance of BC


⇒ BC = √ ((0 – 5)2 + (6 – 0)2)


⇒ BC = √ ((–5)2 + (6)2)


⇒ BC = √ (25 + 36)


⇒ BC = √ 61


We notice that BC = AC = √ 61


∴ Points A, B and C are coordinates of an isosceles triangle.


Question 6.

The distance between the points (0, 8) and (0, –2) is
A. 6

B. 100

C. 36

D. 10


Answer:

Formula used: 


Let the point be A (0, 8) and B (0, –2)


Distance of AB


⇒ AB = √ (0 – 0)2 + (–2 – 8)2)


⇒ AB = √ ((0)2 + (–10)2)


⇒ AB = √ (0 + 100)


⇒ AB = √ 100


⇒ AB = 10


∴ , Option B is correct.


Question 7.

(4, 1), (–2, 1), (7, 1) and (10, 1) are points
A. on x–axis

B. on a line parallel to x–axis

C. on a line parallel to y–axis

D. on y–axis


Answer:


Question 8.

The distance between the points (a, b) and (–a, –b) is
A. 2a

B. 2b

C. 2a + 2b

D. 


Answer:

Formula used: 


Let the point be A (a, b) and B (–a, –b)


Distance of AB


⇒ AB = √ (–a – a)2 + (–b – b)2)


⇒ AB = √ ((–2a)2 + (–2b)2)


⇒ AB = √ (4a2 + 4b2)


⇒ AB = √ 4(a2 + b2)


⇒ AB = 2√ (a2 + b2)


∴ Option D is correct.


Question 9.

The point which is on y–axis with ordinate –5 is
A. (0, −5)

B. (−5, 0)

C. (5, 0)

D. (0, 5)


Answer:

For any point on y–axis, x–coordinate is 0.


∴ the point is (0, –5).


Question 10.

The relation between p and q such that the point (p, q) is equidistant from (–4, 0) and (4, 0) is
A. p = 0

B. q = 0

C. p + q = 0

D. p + q = 8


Answer:

Formula used: 


Let the point be A (p, q), B (–4, 0) and C (4, 0)


Distance of AB


⇒ AB = √ (–4 – p)2 + (0 – q)2)


⇒ AB = √ ((–4 – p)2 + (–q)2)


⇒ AB = √ (16 + p2 + 8p + q2)


Distance of AC


⇒ AC = √ (4 – p)2 + (0 – q)2)


⇒ AC = √ ((4 – p)2 + (–q)2)


⇒ AC = √ (16 + p2 – 8p + q2)


i.e. AB = AC (Given)


⇒ 16 + p2 + 8p + q2 = 16 + p2 – 8p + q2


Squaring both sides


⇒ 16 + p2 + 8p + q2 = 16 + p2 – 8p + q2


⇒ 16 + p2 + 8p + q2 – 16 – p2 + 8p – q2 = 0 …


⇒ 16 p = 0


⇒ p = 0


∴ Option A is correct.