Question

Given the family of lines $a(3x+4y+6)+b(x+y+2)=0.$ The line of the family situated at the greatest distance from the point $P(2,3),$ has the equation

• A. $4x+3y+8=0$
• B. $5x+3y+10=0$
• C. $15x+8y+30=0$
• D. $4x-3y+8=0$

Answer 1

The correct option is A $4x+3y+8=0$

Given the family of lines: $a(3x+4y+6)+b(x+y+2)=0$
So, the family of lines pass through the point of intersection of $3x+4y+6=0$ and $x+y+2=0.$
$\Rightarrow$ Point of intersection $=(-2,0)$


The line situated at the greatest distance from the point $P(2,3)$ is perpendicular to $AP.$
Slope of $AP=\frac{3-0}{2+2}=\frac{3}{4}$
Slope of the required line $=-\frac{4}{3}$
Equation of the required line is,
$y-0=-\frac{4}{3}(x+2)$
$\Rightarrow 4x+3y+8=0$