Question

The family of line x(3a + 2b) + y(a – 2b) = 4a – 3b passes through a point for all values of a and b. Then point will be-

• A. $\left( \frac{1}{2},–1 \right)$
• B. $\left( \frac{17}{8},\frac{5}{8} \right)$
• C. $\left( \frac{5}{8},\frac{- 17}{8} \right)$
• D. None

Answer 1

Answer: (D)

None

Answer 2

Line (3x + y – 4) a + b (2x – 2y + 3) = 0

(3x + y – 4) + $\frac{b}{a}$ (2x – 2y + 3) = 0

L₁ + lL₂ = 0 form

Where l = $\frac{b}{a}$

Ž 3x + y – 4 = 0

2x – 2y + 3 = 0

on solving we get x = 5/8, y = 17/8 point (5/8, 17/8).