CAT 2025

Slot 1

QA

Question & SolutionQues & Sol

The (x, y) coordinates of vertices P, Q and R of a parallelogram PQRS are (-3, -2), (1, -5) and (9, 1), respectively. If the diagonal SQ intersects the x-axis at (a, 0), then the value of a is

1. Concept Used

• Topic: Coordinate Geometry — Properties of a Parallelogram and Equation of a Line
• Formula: In a parallelogram PQRS, the fourth vertex S can be found using the vector addition property:

$$S = P + R - Q$$

The equation of line through two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$$

2. Calculation

We are given $P(-3, -2)$, $Q(1, -5)$, and $R(9, 1)$ as vertices of parallelogram PQRS. The key property of a parallelogram is that its diagonals bisect each other. In parallelogram PQRS, the diagonals are PR and SQ, meaning their midpoints must coincide.

Using the fourth vertex formula:

$$S = P + R - Q = (-3 + 9 - 1,\ -2 + 1 - (-5)) = (5, 4)$$

So $S = (5, 4)$.

Now, the diagonal SQ passes through $S(5, 4)$ and $Q(1, -5)$.

Slope of SQ:

$$m = \frac{-5 - 4}{1 - 5} = \frac{-9}{-4} = \frac{9}{4}$$

Equation of line SQ using point-slope form with point $S(5, 4)$:

$$y - 4 = \frac{9}{4}(x - 5)$$

To find where SQ intersects the x-axis, substitute $y = 0$:

$$0 - 4 = \frac{9}{4}(a - 5)$$

$$-4 = \frac{9}{4}(a - 5)$$

Multiply both sides by 4:

$$-16 = 9(a - 5)$$

$$-16 = 9a - 45$$

$$9a = -16 + 45 = 29$$

$$a = \frac{29}{9}$$

3. Solution

Answer = Option 4 ✅

The diagonal SQ of parallelogram PQRS intersects the x-axis at

$$a = \dfrac{29}{9}$$