Question

If the image of the point $P(4,-13)$ with respect to the line $5x+y+6=0$ is $Q,$ then the area of the triangle formed by the line through $Q,$ which makes equal intercepts on the coordinate axes, with the coordinate axes, can be

• A. $72\text{sq. units}$
• B. $84.5\text{sq. units}$
• C. $112.5\text{sq. units}$
• D. $128\text{sq. units}$

Answer 1

Answer: (B), (C)

The correct options are
B $84.5\text{sq. units}$
C $112.5\text{sq. units}$

Let $Q(h,k)$ be the image of $P(4,-13)$ with respect to the line $5x+y+6=0.$ Then
$\frac{h-4}{5}=\frac{k+13}{1}=\frac{-2(20-13+6)}{25+1}$
$\Rightarrow h=-1,k=-14$
$\Rightarrow Q\equiv (-1,-14)$

Now, equation of a line which makes equal intercepts with coordinate axes, is $x+y=k$ or $x-y=k$

Case $1:$ When $x+y=k$
As it passes through $Q,$ then
$-1-14=k$
$\Rightarrow k=-15$

$\therefore$ Equation of line is $x+y+15=0.$

Hence, area of triangle formed by the above line with the coordinate axes
$=\frac{(15{)}^2}{2}=112.5\text{sq. units}$

Case $2:$ When $x-y=k$
As it passes through $Q,$ then
$-1+14=k$
$\Rightarrow k=13$

$\therefore$ Equation of line is $x-y-13=0.$

Hence, area of triangle formed by the above line with the coordinate axes
$=\frac{(13{)}^2}{2}=84.5\text{sq. units}$