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)

The given system of lines passes through the point of intersection of the straight lines $2x+y-3=0$ and $3x+2y-5=0$ $[L_1+\lambda L_2=0$ form $]$ which is $(1,1)$.
The required line will also pass through this point.
Further, the line will be farthest from point $(4,-3)$
if it is in direction perpendicular to line joining $(1,1)$ and $(4,-3)$.
$\therefore$ The equation of the required line is
$y-1=\frac{-1}{\frac{-3-4}{4-1}}(x-1)$
$\Rightarrow 3x-4y+1=0$