Question

The equation of the line farthest from $(-5,-4)$ belonging to the family of lines $(2+\lambda )x+(3\lambda +1)y+2(2+\lambda )=0$, where $\lambda$ is a variable parameter, is

• A. $3x+4y+6=0$
• B. $3x+4y+3=0$
• C. $4x+3y+3=0$
• D. $4x+3y-3=0$

Answer 1

The correct option is A $3x+4y+6=0$

The given line is $(2x+y+4)+\lambda (x+3y+2)=0$
This always pass through the intersection of $2x+y+4=0$ and
$x+3y+2=0$ i.e., $(-2,0)$
The line through $(-2,0)$ perpendicular to the line joining the points $(-5,-4)$ and $(-2,0)$ is the required line.
$\therefore$ its equation is $y-0=-\frac{3}{4}(x+2)$ i.e .,$3x+4y+6=0.$