CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 27

Answer: (a) (−1, 0)

Explanation:
1. The diagonals of a parallelogram bisect each other, so the midpoint of the given diagonal is ((3 − 6)/2, (−4 + 5)/2) = (−3/2, 1/2).
2. Let the fourth vertex be (x, y); it forms the other diagonal with (−2, 1): (x − 2)/2 = −3/2 and (y + 1)/2 = 1/2.
3. x = −1 and y = 0, so the vertex is (−1, 0).

Teacher's Note:
1. Equal midpoints of the diagonals identify the parallelogram's missing vertex.
2. Fourth vertex = sum of the opposite pair − the known vertex.

Answer: (d) 1

Explanation:
1. PR/PQ = 3/5 means PR : RQ = 3 : 2.
2. x = (3 × (−1) + 2 × 4)/(3 + 2) = 5/5 = 1.

Teacher's Note:
1. Convert the ratio PR/PQ to PR : RQ before using the section formula.
2. The part-to-whole ratio 3/5 gives 3 : 2.

Answer: (d) AP = ½ AB

Explanation:
1. The midpoint of AB is ((4 + 8)/2, (2 + 4)/2) = (6, 3), which is P.
2. So P is the midpoint and AP = ½ AB.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (b) (0, −1)

Explanation:
1. The diagonals AC and BD bisect each other. Midpoint of AC = (3, 3).
2. For BD: (6 + x)/2 = 3 and (7 + y)/2 = 3, so x = 0 and y = −1.
3. D = (0, −1).

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (b) −12

Explanation:
1. x-coordinate of the midpoint = (−6 + (−2))/2 = −4.
2. a/3 = −4, so a = −12.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (c) 3abc

Explanation:
1. Centroid: ((a + b + c)/3, (b + c + a)/3) = (0, 0), so a + b + c = 0.
2. Using the identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) = 0, we get a³ + b³ + c³ = 3abc.

Teacher's Note:
1. The identity for the sum of cubes is useful whenever a + b + c = 0.
2. Both coordinates of the centroid give the same condition.

Answer: (c) 3 : 2

Explanation:
1. Let the ratio be k : 1 so that y = (8k + 3)/(k + 1) = 6.
2. 8k + 3 = 6k + 6, so 2k = 3 and k = 3/2, that is 3 : 2.
3. Check with x: (3 × 2 + 2 × (−4))/5 = −2/5.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (a) (0, 0)

Explanation:
1. Points on the perpendicular bisector are equidistant from A and B.
2. For (0, 0): distance to A = √(4 + 25) = √29 and to B = √(4 + 25) = √29.
3. Other options give unequal distances, e.g. (0, 2): √(4 + 49) and √(4 + 9).

Teacher's Note:
1. The midpoint of a segment always lies on its perpendicular bisector.
2. Here (0, 0) is the midpoint.

Answer: (b) 3 : 1

Explanation:
1. On the y-axis x = 0. Let the ratio be k : 1.
2. x = (k × 1 + 1 × (−3))/(k + 1) = 0, so k = 3.
3. The ratio is 3 : 1.

Teacher's Note:
1. The y-axis divides a segment in the ratio of the absolute x-coordinates: 3 : 1.
2. Section formula with x = 0.

Answer: (d) IV quadrant

Explanation:
1. x = (1 × 3 + 2 × 7)/3 = 17/3 and y = (1 × 3 + 2 × (−6))/3 = −3.
2. The point (17/3, −3) has positive x and negative y, so it is in the IV quadrant.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (b) 5

Explanation:
1. Area of the triangle = 0: 5(p − 2) + 1(2 − 1) + 4(1 − p) = 0.
2. 5p − 10 + 1 + 4 − 4p = 0, so p = 5.

Teacher's Note:
1. Collinear points form a triangle of zero area.
2. Check: the slope between (5, 1) and (4, 2) is −1, and between (1, 5) and (4, 2) is −1.

Answer: (a) a²

Explanation:
1. Area = ½ |a(c − (c − a)) + a((c − a) − (c + a)) + (−a)((c + a) − c)|.
2. = ½ |a² − 2a² − a²| = ½ × 2a² = a².

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (c) \( \frac{11}{2} \)

Explanation:
1. Shoelace formula: sum of xᵢyᵢ₊₁ − xᵢ₊₁yᵢ = (2 − 12) + (18 − 10) + (20 − 9) + (6 − 4) = −10 + 8 + 11 + 2 = 11.
2. Area = ½ × 11 = 11/2.

Teacher's Note:
1. List the vertices in order (clockwise or anticlockwise).
2. Alternatively split the quadrilateral into two triangles.

Answer: (c) 2a = b

Explanation:
1. The slope of OA = 2/1 = 2, and the slope of OC = b/a.
2. Collinear points have equal slopes: b/a = 2, so b = 2a.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (d) \( \frac{-1}{3} \)

Explanation:
1. Area = 0: k(3k − 1) + 3k(1 − 2k) + 3(2k − 3k) = 0.
2. 3k² − k + 3k − 6k² − 3k = 0, which gives −3k² − k = 0.
3. k(3k + 1) = 0, so k = −1/3 since k ≠ 0.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (a) 13 units

Explanation:
1. AB = √((−2 + 7)² + (−5 − 7)²) = √(25 + 144) = √169 = 13.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (a) −1

Explanation:
1. The line through (0, 4) and (4, 0) is x + y = 4.
2. (6, 2p) lies on it: 6 + 2p = 4, so p = −1.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (b) (0, −2)

Explanation:
1. Let the point be (0, y). Then 25 + (y + 2)² = 9 + (y − 2)².
2. 25 + y² + 4y + 4 = 9 + y² − 4y + 4, so 8y = −16 and y = −2.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (b) (8, −3) and (−2, 1)

Explanation:
1. The centre is the midpoint of the diagonal: (3, −1). The half-diagonal vector from the centre to (5, 4) is (2, 5).
2. Rotating this by 90° gives (−5, 2) and (5, −2). So the other vertices are (3 − 5, −1 + 2) = (−2, 1) and (3 + 5, −1 − 2) = (8, −3).

Teacher's Note:
1. The diagonals of a square are equal, perpendicular and bisect each other.
2. Check: distance between (−2, 1) and (8, −3) equals the other diagonal √116.

Answer: (c) −1

Explanation:
1. Slope between (−5, 1) and (4, −2) = −3/9 = −1/3.
2. Slope between (−5, 1) and (1, p) = (p − 1)/6 = −1/3, so p − 1 = −2 and p = −1.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (c) √85

Explanation:
1. The midpoint of AB is D(−1, −1).
2. CD = √((5 + 1)² + (−8 + 1)²) = √(36 + 49) = √85.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (b) (−56/9, −3)

Explanation:
1. x = (7 × (−6) + 2 × (−7))/9 = −56/9.
2. y = (7 × (−5) + 2 × 4)/9 = −27/9 = −3.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (b) (9, 13)

Explanation:
1. The centre is the midpoint of the diameter: (5 + x)/2 = 7 and (−7 + y)/2 = 3.
2. x = 9 and y = 13.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.

Answer: (a) 3 : 4

Explanation:
1. Let the ratio be m : 1; the point is ((2m + 1)/(m + 1), (7m + 3)/(m + 1)).
2. Substitute in the line: 3(2m + 1) + (7m + 3) − 9(m + 1) = 0, so 6m + 3 + 7m + 3 − 9m − 9 = 0.
3. 4m = 3, so m = 3/4 and the ratio is 3 : 4.

Teacher's Note:
1. Write the formula first and substitute carefully with signs.
2. Verify the result by checking the point or ratio you found.