Question

The locus of $P$ such that the area of $\Delta PAB=12$ sq. units, where $A=(2,3)$ and $B(-4,5)$ is/are

• A. $x+3y=23$
• B. $x+3y+23=0$
• C. $x+3y=1$
• D. $x+3y+1=0$

Answer 1

Answer: (A), (D)

The correct options are
A $x+3y=23$
D $x+3y+1=0$

Equation of the line passing through $A$ and $B$ is
$y-5=\frac{5-3}{-4-2}(x+4)\Rightarrow x+3y=11$
and $|AB|=2\sqrt{10}$
Let $P$ be $(x,y)$.
So the perpendicular distance from $P$ to line $x+3y=11$
$d=\frac{|x+3y-11|}{\sqrt{10}}$
So area of $△PAB=\frac{1}{2}\cdot \frac{|x+3y-11|}{\sqrt{10}}\cdot 2\sqrt{10}\Rightarrow 12=|x+3y-11|$
$\therefore$ locus of $P$ will be
$x+3y=23$ or $x+3y+1=0$


Alternate solution:
Let $P(x,y)$
so area of the triangle will be
$\frac{1}{2}|x(3-5)+2(5-y)-4(y-3)|=12\Rightarrow |-2x-6y+22|=24\Rightarrow x+3y=23$
or $x+3y+1=0$