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Question 63

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

Solution

Given (P(-3,-2)), (Q(1,-5)), and (R(9,1)).

For parallelogram (PQRS), S = P + R - Q = (-3,-2) + (9,1) - (1,-5) = (5,4)

Diagonal SQ passes through S(5,4) and Q(1,-5).

Slope m = $$\dfrac{-5-4}{1-5} = \dfrac{9}{4}$$

Equation of SQ is $$y - 4 = \dfrac{9}{4}(x - 5)$$

At the x-axis, y = 0: $$-4 = \dfrac{9}{4}(a - 5)$$

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

$$9a = 29$$

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

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

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