Question

Let $\Delta ABC$ is an isosceles triangle with $AB=AC.$ If $B=\left(0,a\right),C=\left(2a,0\right)$ and the equation of $AB$ is $3x-4y+4a=0$ , then the equation of side $AC$ is

• A. $y=8x-16a$
• B. $3y=4x-8a$
• C. $x=2a$
• D. $y+8x=16a$

Answer 1

Answer: (C)

Let, $\angle ABC=\angle ACB=\theta$ and slope of $AC=m$
Slope of $BC$ $=-\frac{1}{2}$
Slope of $AB=\frac{3}{4}$
Now, tan $\left(\angle ABC\right)=tan\left(\angle ACB\right)$
$\Rightarrow$ $\frac{\frac{3}{4}-\left(-\frac{1}{2}\right)}{1+\left(\frac{3}{4}\right)\left(-\frac{1}{2}\right)}=\frac{-\frac{1}{2}-m}{1+\left(-\frac{1}{2}m\right)}$
$\Rightarrow$ $2=\frac{-\frac{1}{2}-m}{1-\frac{m}{2}}\Rightarrow 2-m=-\frac{1}{2}-m\Rightarrow m=$ not defined
$\Rightarrow$ equation of $AC$ is $x=2a$