Question

If $x-2y+4=0$ and $2x+y-5=0$ are the sides of an isosceles triangle having area $10$ sq. units the equation of third side is

• A. $3x-y=-9$
• B. $3x-y-11=0$
• C. $x-3y=19$
• D. $3x-y+15=0$

Answer 1

Answer: (A), (B)

The correct options are
A $3x-y=-9$
B $3x-y-11=0$

Given sides

$x-2y+4=0$-----(1) and

$2x+y-5=0$-----(2)

from eq (1) and (2) we get intersecting point

$I(\frac{6}{5},\frac{13}{5})$

Eq of angle bisector of given lines

$x-2y+4=\pm (2x+y-5)$

on solving we get

$x+3y=9$ and $3x-y=1$

Side BC will be parallel to these bisectors.Let

$AD=a$ where D is foot of perpendicular

$\Rightarrow AB=a\sqrt{2}$ and area of $\Delta ABC$ is

$\frac{1}{2}\times (a\sqrt{2}{)}^2=a^2=10$

$\Rightarrow a=\sqrt{10}$

Let eq of BC be $x+3y=\lambda$

$\sqrt{10}=\frac{\frac{6}{5}-\frac{39}{5}-\lambda }{\sqrt{10}}\Rightarrow \lambda =-1,19$

Eq of BC is $x+3y=-1$ or $x+3y=19$

If eq of BC is $3x-y=\lambda$

$\sqrt{10}=\frac{\frac{18}{5}-\frac{13}{5}-\lambda }{\sqrt{10}}\Rightarrow \lambda =-9,11$

Eq of BC is $3x-y=-9$ or $3x-y=11$