Question

Given the system of straight lines a(2x + y – 3) + b (3x + 2y – 5) = 0, the line of the system situated farthest from the point (4, –3) has the equation –

• A. 4x + 11y – 15 = 0
• B. 7x + y – 8 = 0
• C. 4x + 3y – 7 = 0
• D. 3x – 4y + 1 = 0

Answer 1

Answer: (D)

3x – 4y + 1 = 0

Answer 2

The system of lines (given) pass through (1, 1).

So, the required line is the one through (1, 1) and

perpendicular to the join of (1, 1) and (4, –3).

\ Its equation is $\frac{y - 1}{x - 1}$ = $\frac{3}{4}$ i.e. 3x – 4y + 1 = 0