Question

Straight lines $x–y=7$ and $x+4y=2$ intersect at $B$. Points $A$ and $C$ are so chosen on these two lines such that $AB=AC$. The equation of line $AC$ passing through $(2,–7)$ is

• A. $x–y–9=0$
• B. $23x+7y+3=0$
• C. $2x–y–11=0$
• D. $7x–6y–56=0$

Answer 1

The correct option is B $23x+7y+3=0$

Let the slope of the line $AC$ be $m$.
Then, the equation of $AC$ be $y+7=m(x-2)$.

Slope of the line $x-y=7$ is $1$ and slope of the line $x+4y=2$ is $-\frac{1}{4}$.

Let the angle between line $AB$ and $BC$ be $\theta$.
$\therefore \tan \theta =\left|\frac{1-\left(\frac{-1}{4}\right)}{1+1\left(\frac{-1}{4}\right)}\right|$
$\Rightarrow \tan \theta =\frac{5}{3}$

In $△ABC,AB=AC$
$\Rightarrow \angle ABC=\angle ACB=\theta$
$\Rightarrow \tan \left(\angle ABC\right)=\tan \left(\angle ACB\right)$
$\Rightarrow \frac{5}{3}=\left|\frac{m-\left(\frac{-1}{4}\right)}{1+m\left(\frac{-1}{4}\right)}\right|$
$\Rightarrow \frac{5}{3}=\pm \left(\frac{m+\frac{1}{4}}{1-\frac{m}{4}}\right)$

On solving, we get
$m=1,-\frac{23}{7}$
$m\ne 1$ because if $m=1$, then we get the line which is parallel to $x-y=7$
So, $m=\frac{-23}{7}$

$\therefore$ Equation of line $AC$ is $23x+7y+3=0$