Question

In an isosceles triangle $ABC$, the coordinates of the point $B$ and $C$ on the base $BC$ are respectively $(2,1)$ and $(1,2)$. If the equation of the line $AB$ is $y=\frac{x}{2},$ then equation of the line $AC$ is

• A. $2y=x+3$
• B. $y=x$
• C. $y=\frac{1}{2}\left(x-1\right)$
• D. $y=x-1$

Answer 1

The correct option is B $y=x$

As we know the $\perp$ bisector of base of an isosceles triangle will always pass through third vector.

Let

Mid point of BC is

$D=\left[\frac{2+1}{2};\frac{1+2}{2}\right]=\left(\frac{3}{2},\frac{3}{2}\right)$

Slope

$\begin{array}{c}{m}_1=\frac{y_2-y_1}{x_2-x_1}=\frac{2-1}{1-2}=-1\\ m_1\cdot m_2=-1\\ \therefore m_2=1\end{array}$

Hence, equation becomes

$\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-\frac{3}{2}=1\left(x-\frac{3}{2}\right)\Rightarrow y=x\end{array}$