Question

Equation of straight line ax + by + c = 0,

where 3a + 4b + c = 0, which is at maximum distance from

(1, – 2) is –

• A. 3x + y – 17 = 0
• B. 4x + 3y – 24 = 0
• C. 3x + 4y – 25 = 0
• D. x + 3y – 15 = 0

Answer 1

Answer: (D)

x + 3y – 15 = 0

Answer 2

It passes through a fixed point (3, 4)

Slope of line joining (3, 4) and (1, – 2) is $\frac{- 6}{- 2}$ = 3

\ Slope of required line = – $\frac{1}{3}$

equation is y – 4 = – $\frac{1}{3}$ (x – 3)

x + 3y – 15 = 0