### Coordinate Geometry Class 9th Mathematics Term 1 Tamilnadu Board Solution

##### Question 1.The point (–2, 7) lies is the quadrantA. IB. IIC. IIID. IVAnswer: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 onA. OXB. OYC. 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 IIIA. a > 0, b < 0B. a < 0, b < 0C. a > 0, b > 0D. a < 0, b > 0Answer: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) isA. 2B. 4C. √2D. 8Answer: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) isA. an isosceles triangleB. right triangleC. scalene triangleD. an equilateral triangleAnswer: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 = 10Distance of AC⇒ AC = √ ((0 – (–5))2 + (6 – 0)2)⇒ AC = √ ((5)2 + (6)2)⇒ AC = √ (25 + 36)⇒ AC = √ 61Distance of BC⇒ BC = √ ((0 – 5)2 + (6 – 0)2)⇒ BC = √ ((–5)2 + (6)2)⇒ BC = √ (25 + 36)⇒ BC = √ 61We 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) isA. 6B. 100C. 36D. 10Answer: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 pointsA. on x–axisB. on a line parallel to x–axisC. on a line parallel to y–axisD. on y–axisAnswer: Question 8.The distance between the points (a, b) and (–a, –b) isA. 2aB. 2bC. 2a + 2bD. 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 isA. (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) isA. p = 0B. q = 0C. p + q = 0D. p + q = 8Answer: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 + q2Squaring 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.

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