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

System of lines (given) pass through (1, 1). So the required line is passes through (1, 1) and perpendicular to the line joining (1, 1) and (4, –3)

So eqⁿ is

$\frac { y - 1 } { x - 1 }$ = $\frac { 3 } { 4 }$ Ž 3x – 4y + 1 = 0