Question

The vertices of a triangle are A(-1, -7), B(5,1) and C(1,4). Find the equation of the bisector of < ABC.

• A. x+7y-2=0
• B. x-7y-2=0
• C. x-7y+2=0
• D. None of these

Answer 1

The correct option is A x+7y-2=0

Given that, Vertices are $A(-1,-7)$ , $B(5,1)$ , $C(1,4)$

We know that, angular bisector divides third side in the ration of two sides, i.e. $AP:PC=AB:BC$

Here, $AB=\sqrt{(5+1{)}^2+(7+1{)}^2}=\sqrt{36+64}=10BC=\sqrt{(5-1{)}^2+(1-4{)}^2}=\sqrt{16+9}=5$

$\Rightarrow$ P divides AC in $10:5$ ratio i.e $2:1$

we know that, if P divides AC in $m:n$ ratio, then

$P=(\frac{m{x}_1+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

$\Rightarrow$ Here, $P=[\frac{2\left(1\right)+1(-1)}{2+1},\frac{2\left(4\right)+1(-7)}{2+1}]=(\frac{2-1}{3},\frac{8-7}{3})$

$\Rightarrow P=(\frac{1}{3},\frac{1}{3})$

We have coordinates of $B(5,1)$ and $P(\frac{1}{3},\frac{1}{3})$

So, equation is $y-1=\frac{\frac{1}{3}-1}{\frac{1}{3}-5}(x-5)\Rightarrow y-1=\frac{\frac{-2}{3}}{\frac{-14}{3}}(x-5)\Rightarrow 7y-7=x-5$

$\therefore x-7y+2=0$ is the required equation.