CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 17

Answer: (c) (−3, −5)

Explanation:
1. Reflection in the x-axis changes the sign of the y-coordinate only.
2. So (−3, 5) becomes (−3, −5).

Teacher's Note:
1. Reflection in x-axis: y changes sign.
2. Reflection in y-axis: x changes sign.

Answer: (d) 1

Explanation:
1. By the section formula, the y-coordinate of P is \( \frac{3y+1\times5}{3+1} \).
2. So \( \frac{3y+5}{4}=2 \), 3y = 3 and y = 1.

Teacher's Note:
1. Use the section formula with m : n = 3 : 1.
2. The x-coordinates are already equal, so only y is needed.

Answer: (b) 1 : 6

Explanation:
1. Let the ratio be k : 1. Then \( \frac{4k-3}{k+1} \) = −2, so 6k = 1 and k = 1/6.
2. The ratio is 1 : 6. Check on y: \( \frac{-9k+5}{k+1} \) = 3.

Teacher's Note:
1. Use the section formula on the x-coordinate.
2. Verify with the y-coordinate.

Answer: (b) 3 : 4

Explanation:
1. Any point on the y-axis has x = 0. Let the ratio be k : 1: \( \frac{-4k+3}{k+1} \) = 0.
2. So k = 3/4, and the ratio is 3 : 4.

Teacher's Note:
1. On the y-axis the x-coordinate is 0.
2. Use the section formula.

Answer: (a) (2, 0)

Explanation:
1. Let the point be (x, 0). Then (x − 5)² + 16 = (x + 2)² + 9.
2. −10x + 41 = 4x + 13, so x = 2.

Teacher's Note:
1. A point on the x-axis has y = 0.
2. Equate the squared distances.

Answer: (c) \( \left(\frac{1}{3},-\frac{34}{3}\right) \)

Explanation:
1. The point nearer to (−2, −19) divides the segment in the ratio 1 : 2.
2. x = \( \frac{1\times5+2\times(-2)}{3} \) = 1/3 and y = \( \frac{1\times4+2\times(-19)}{3} \) = −34/3.

Teacher's Note:
1. Trisection points use the ratios 1 : 2 and 2 : 1.
2. Apply the section formula.

Answer: (b) −15

Explanation:
1. The mid-point is \( \left(\frac{3+k}{2},\frac{11}{2}\right) \).
2. Substitute: (3 + k) + 11 + 1 = 0, so k = −15.

Teacher's Note:
1. Use the mid-point formula.
2. Put the point on the line.

Answer: (b) −12

Explanation:
1. x-coordinate of the mid-point = (−6 − 2)/2 = −4.
2. So a/3 = −4 and a = −12.

Teacher's Note:
1. Use the mid-point formula for x.
2. The y-coordinate (5 + 3)/2 = 4 checks out.

Answer: (c) −3

Explanation:
1. PQ² = (x − 2)² + (−7 − 5)² = 169, so (x − 2)² = 25.
2. x = 7 or x = −3; only −3 is among the options.

Teacher's Note:
1. Use the distance formula.
2. Two values are possible; check the options.

Answer: (a) \( \left(5,\frac{11}{2}\right) \)

Explanation:
1. PR = √(8² + 6²) = 10, so PQ : QR = 2.5 : 7.5 = 1 : 3.
2. Q = \( \left(\frac{1\times11+3\times3}{4},\frac{1\times10+3\times4}{4}\right)=\left(5,\frac{11}{2}\right) \).

Teacher's Note:
1. Find the full length PR first.
2. Then use the section formula.

Answer: (c) 3 : 2, y = 3

Explanation:
1. x-coordinate: \( \frac{6k-4}{k+1} \) = 2, so 4k = 6 and k = 3/2. The ratio is 3 : 2.
2. y = (3k + 3)/(k + 1) = 3.

Teacher's Note:
1. Use the section formula on x to get the ratio.
2. The y-coordinates are equal, so y = 3.

Answer: (c) 4 : 9

Explanation:
1. Put 3x + 4y − 7 at the end points: at (1, 2) it is 4 and at (−2, 1) it is −9.
2. The ratio is 4 : 9 (internal, since the signs differ).

Teacher's Note:
1. The ratio equals the ratio of the absolute values at the two points.
2. Opposite signs mean internal division.

Answer: (b) \( \left(\frac{1}{3},\frac{5}{3}\right) \)

Explanation:
1. x = \( \frac{1\times(-1)+2\times1}{3} \) = 1/3 and y = \( \frac{1\times1+2\times2}{3} \) = 5/3.

Teacher's Note:
1. Use the section formula with m : n = 1 : 2.
2. The point lies nearer to A.

Answer: (b) (0, 3)

Explanation:
1. x = (3 − 8 + 5)/3 = 0 and y = (−7 + 6 + 10)/3 = 3.
2. Centroid = (0, 3).

Teacher's Note:
1. The centroid is the mean of the vertices.
2. Add the x's and y's separately.