Coordinate Geometry Class 10 Previous Year Questions (Chapter 7)

1. 1

   Step 1: Mid-point of diagonal AC

   1 mark (suggested)

   The diagonals of a parallelogram bisect each other, so the mid-point of AC equals the mid-point of BD. For AC, the x-coordinate is (4+4)/2 = 4 and the y-coordinate is (5+3)/2 = 4.

   Why? The diagonals of a parallelogram bisect each other, so their mid-points are the same point. This turns the geometry into two coordinate equations.

2. 2

   Step 2: Equate mid-point of BD and solve

   1 mark (suggested)

   For BD, the x-coordinate is (m+1)/2 and the y-coordinate is (6+n)/2. Equating x-coordinates: (m+1)/2 = 4, so m = 7. Equating y-coordinates: (6+n)/2 = 4, so n = 2. Therefore, m = 7 and n = 2.

   Why? Two points are equal only when their x-coordinates are equal and their y-coordinates are equal. Solving each linear equation gives one unknown.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Show that B is the mid-point of AC

   1 mark (suggested)

   Let C = (x, y). C lies on AB produced beyond B, so A, B and C are collinear with B between A and C, and AC = AB + BC. Given AC = 2BC, AB + BC = 2BC, so AB = BC. Hence B is the mid-point of AC.

   Why? On a straight line, the whole distance is the sum of the two adjacent distances. Equal adjacent distances mean the common point is the mid-point.

2. 2

   Step 2: Use the mid-point formula

   1 mark (suggested)

   Since B is the mid-point of AC, C = 2B - A. So x = 2(4) - (-1) = 9 and y = 2(-3) - 7 = -13. Therefore the required point is C(9, -13).

   Why? The mid-point formula says the coordinates of the mid-point are the averages of the coordinates of the endpoints. Rearranging gives C = 2B - A, so each coordinate of C is twice the corresponding coordinate of B minus the coordinate of A.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Let the ratio be k : 1

   1 mark (suggested)

   Let the point (3, y) divide AB in the ratio k : 1. Using the section formula, 3 = (6k − 2)/(k + 1). Solving, 6k − 2 = 3k + 3, so k = 5/3.

   Why? The section formula connects an unknown ratio to the coordinates of the dividing point; using the given x-coordinate creates the equation that fixes the ratio.

2. 2

   Step 2: Find y and state the ratio

   1 mark (suggested)

   Using the section formula for y, y = (3k − 5)/(k + 1) = (3(5/3) − 5)/((5/3) + 1) = 0. Hence the ratio is 5 : 3 and y = 0.

   Why? With the ratio known, the section formula gives the other coordinate directly.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)

1. 1

   Step 1: Apply distance formula to PQ

   1 mark

   PQ = √((6 − 1)² + (−1 − 3)²) = √(5² + (−4)²) = √41.

   Why? The distance formula gives the length of the segment joining two points from their coordinates. Since the differences are squared, subtracting in either order gives the same distance.

2. 2

   Step 2: Apply distance formula to QR

   ½ mark

   QR = √((x − 1)² + (8 − 3)²) = √((x − 1)² + 5²) = √((x − 1)² + 25).

   Why? QR also comes from the distance formula. It is written with the unknown x because R's x-coordinate is x.

3. 3

   Step 3: Solve PQ = QR

   ½ mark

   PQ = QR gives √41 = √((x − 1)² + 25), so 41 = (x − 1)² + 25, i.e. (x − 1)² = 16. Hence x − 1 = ±4, so x = 5 or x = −3.

   Why? If two non-negative numbers are equal, their squares are equal. This removes the square roots and gives a simple equation in x.

Marks: 1 + ½ + ½ = 2 · Split as in the CBSE marking scheme, 2023 · Basic 430/2/1 Q23

1. 1

   Step 1: Let the ratio be k : 1

   1 mark (suggested)

   Let the ratio be k : 1. By the section formula, the point of division is ((k×1 + 1×6)/(k+1), (k×6 + 1×(−3))/(k+1)).

   Why? The section formula gives the coordinates of the point that divides a segment internally in the ratio k : 1.

2. 2

   Step 2: Solve for k

   1 mark (suggested)

   Since the point lies on y = x, (k+6)/(k+1) = (6k−3)/(k+1), so k = 9/5. Hence the required ratio is 9 : 5.

   Why? A point lies on y = x when its x-coordinate equals its y-coordinate; equating them determines k.

Marks: 1 + 1 = 2 · Suggested split (CBSE’s usual pattern for 2-mark questions)