Question

The point P(a, b) lies on the straight line 3x + 2y = 13 and the point Q (b, a) lies on the straight line 4x – y = 5, then the equation of line PQ is

• A. x – y = 5
• B. x + y = 5
• C. x + y = – 5
• D. x – y = – 5

Answer 1

Answer: (B)

x + y = 5

Answer 2

Point P (a, b) lies on 3x + 2y = 13

So, 3a + 2b = 13 ….(i)

Point Q (b, a) lies on 4x – y = 5

So, 4b – a = 5 ….(ii)

By Solving (i) & (ii) Ž a = 3 & b = 2

\ P (a, b) = (3, 2) & Q (b, a) = (2, 3)

Now, equation of PQ is y – y₁ = (x –x₁)

Ž x + y = 5