CBSE Class 8 Mathematics Coordinate Geometry MCQs Set 16

Answer: (a) 5, −3

Explanation:
1. PQ² = (6 − 1)² + (−1 − 3)² = 25 + 16 = 41.
2. QR² = (x − 1)² + (3 − 8)² = (x − 1)² + 25 = 41, so (x − 1)² = 16 and x − 1 = ±4.
3. x = 5 or x = −3.

Teacher's Note:
1. Use the distance formula and equate the squares of the distances.
2. Both roots are valid.

Answer: (a) a = −3, b = 1

Explanation:
1. The diagonals of a square bisect each other. The diagonal from (0, −2) to (0, 4) has mid-point (0, 1).
2. The other diagonal joins (3, 1) and (a, b), so (3 + a)/2 = 0 and (1 + b)/2 = 1.
3. a = −3 and b = 1.

Teacher's Note:
1. Use the mid-point of a diagonal.
2. Check: side = 3√2 for all four sides.

Answer: (d) Point Q(−5, 4) lies in the second quadrant

Explanation:
1. P(5, 0) lies on the x-axis, R(−3, −3) is in the third quadrant and S(−1, 1) is in the second quadrant.
2. Q(−5, 4) has negative x and positive y, so it is in the second quadrant.

Teacher's Note:
1. Quadrants: I (+, +), II (−, +), III (−, −), IV (+, −).
2. Points on the axes belong to no quadrant.

Answer: (a) 0

Explanation:
1. Every point on the x-axis has y-coordinate 0, so the ordinate is 0.

Teacher's Note:
1. Points on the x-axis have the form (x, 0).
2. Ordinate = y-coordinate.

Answer: (d) 13

Explanation:
1. Distance = √((3 + 2)² + (7 + 5)²) = √(25 + 144) = √169 = 13.

Teacher's Note:
1. Apply the distance formula.
2. 5-12-13 triple.

Answer: (a) (3, 0)

Explanation:
1. Let the point be (x, 0): (x − 7)² + 36 = (x + 3)² + 16.
2. −14x + 85 = 6x + 25, so x = 3.

Teacher's Note:
1. A point on the x-axis has y = 0.
2. Check: both distances are √52.

Answer: (c) neither right-angled, isosceles nor equilateral

Explanation:
1. Sides: 3, 5 and √10.
2. No two sides are equal, and 3² + (√10)² = 19 ≠ 25, so it is not right-angled.

Teacher's Note:
1. Compute all three side lengths.
2. Check Pythagoras for every pairing.

Answer: (c) 2, −4

Explanation:
1. PA · PB = 0: (5 − 12)(11 − 12) + (3 − y)(−5 − y) = 0.
2. 7 + (y² + 2y − 15) = 0, so y² + 2y − 8 = 0 and y = 2 or −4.

Teacher's Note:
1. Perpendicular sides have dot product zero.
2. (y − 2)(y + 4) = 0.

Answer: (b) 2

Explanation:
1. (x + 2)² + 25 = (x − 2)² + 9, so 8x = −16 and x = −2.
2. −x = 2.

Teacher's Note:
1. Equate squared distances; x² terms cancel.
2. Check: both distances are √41.

Answer: (c) 6

Explanation:
1. Diagonals: from (3, 0) to (−1, 4) has length 4√2 and from (4, 5) to (−2, −1) has length 6√2.
2. Area = (1/2) × 4√2 × 6√2 = 24, and one-fourth is 6.

Teacher's Note:
1. Area of a rhombus = half the product of the diagonals.
2. The shoelace formula also gives 24.

Answer: (c) (4, 5)

Explanation:
1. The centre is the mid-point of the diameter: ((1 + 7)/2, (1 + 9)/2) = (4, 5).

Teacher's Note:
1. Centre = mid-point of any diameter.
2. Radius would be 5 units.

Answer: (c) 24

Explanation:
1. Diagonals have lengths 4√2 and 6√2.
2. Area = (1/2) × 4√2 × 6√2 = 24 square units.

Teacher's Note:
1. Area = (1/2) d₁ d₂.
2. Verify with the shoelace formula.

Answer: (d) 4

Explanation:
1. The slope between the first two points is (1 + 2)/(5 − 7) = −3/2.
2. Slope between (5, 1) and (3, K) is (K − 1)/(3 − 5) = −3/2, so K − 1 = 3 and K = 4.

Teacher's Note:
1. Equal slopes mean collinear points.
2. Area of the triangle would be zero.