Question

If $x–2y+4=0$ and $2x+y–5=0$ are the sides of a isosceles triangle having area $10$ sq unit. Equation of third side is

• A. x + 3y = 19
• B. 3x – y – 11 = 0
• C. x + 3y = 19
• D. 3x – y + 15 0

Answer 1

The correct option is B 3x – y – 11 = 0

Given sides of a isosceles triangle are $x–2y+4=0$ and $2x+y–5=0$ with area $10$ sq. unit
Consider $l_1:x–2y+4=0$.....(i)
$\Rightarrow 2y=x+4$
$\therefore y=\frac{x}{2}+2$ is in the form of $y=mx+c$
Slope $m=m_1=\frac{1}{2}$
Similar way, for $l_2:2x+y–5=0$.....(ii)
$\Rightarrow y=-2x+5$
Slope $m_2=-2$.
Thus, $m_1\times m_2=-1$
So, both the lines intersects perpendicularly at the point $(\frac{6}{5},\frac{13}{5})$.

Euqation of angle bisector of given lines: $x-2y+4=\pm (2x+y-5)$
$\Rightarrow x-2y+4=(2x+y-5)$ and $x-2y+4=-(2x+y-5)$
$\Rightarrow x-2y+4-2x-y+5=0$ and $x-2y+4+2x+y-5=0$
$\therefore x+3y=9$ and $3x-y=1$

Side BC will be parallel to these bisectors.
Let AD=a where D is foot of perpendicular.
$\Rightarrow AB=a\sqrt{2}$
Area of $\Delta ABC$ is $=\frac{1}{2}\times 2a\times a=10$
$\Rightarrow a^2=10$
$\therefore a=\sqrt{10}$
Case (1): Let the equation of BC be $x+3y=c$
Since, the distance from A to line BC is $\sqrt{10}$.
$\Rightarrow \frac{|\frac{6}{5}+3\left(\frac{13}{5}\right)-c|}{\sqrt{1^2+3^2}}=\sqrt{10}$
$\Rightarrow \frac{|\frac{6+39-5c}{5}|}{\sqrt{1+9}}=\sqrt{10}$
$\Rightarrow \frac{45-5c}{5}=\pm 10$
$\Rightarrow 45-5c=\pm 50$
$\Rightarrow 5c=45-50$ or $5c=-45-50$
$\Rightarrow 5c=-5$ or $5c=-95$
$\therefore c=-1$ or $c=-19$
Hence, the equation of BC is $x+3y=-1$ or $x+3y=-19$