Question

Consider a triangle $ABC$ with vertex $A (2,-4)$. The internal bisectors of the angles $B$ and $C$ are $x+y=2$ and $x-3 y=6$ respectively. Let the two bisectors meet at $I$.
If $\left( x _1, y _1\right)$ and $\left( x _2, y _2\right)$ are the coordinates of the point $B$ and $C$ respectively, then the value of $\left(x_1 x_2+y_1 y_2\right)$ is equal to

• A. 4
• B. 5
• C. 6
• D. 8

Answer 1

Answer: (D)

Image of the point $A (2,-4)$ in the internal angular bisector $BP$ lies on $BC$. Line perpendicular to $x + y =2$ is $x-y=\lambda$. It passes through $(2,-4) \Rightarrow \lambda=6$
$\therefore x-y=6$
Solving (1) and (2), we get $M(4,-2)$
Hence image of $A$ i.e. $A ^{\prime}(6,0)$
Again a line perpendicular to $x-3 y=6$ is $3 x+y=k$, passing through $(2,-4)$
$\therefore 6-4=k \Rightarrow k=2 $
$\therefore 3 x+y=2$
Solving (3)and (4), we get $N=\left(\frac{6}{5}, \frac{-8}{5}\right)$

Hence using mid-point formula, image of $A$ is $x -3 y =6$ is $A ^{\prime \prime}\left(\frac{2}{5}, \frac{4}{5}\right)$ which lies on $BC$.
$\therefore $ Equation of $BC$ is the line passing through $\left(\frac{2}{5}, \frac{4}{5}\right)$ and $(6,0)$, is
$(y-0)=\frac{\frac{-4}{5}}{6-\frac{2}{5}}(x-6) \Rightarrow y=\frac{-1}{7}(x-6) \Rightarrow x+7 y=6$
Also $x -3 y =6$ bisector of $\angle C$.
Solving $y=0$ and $x=6$, we get $C(6,0)$
||lly solving $x+7 y=6$ with $x+y=2$, we get $B\left(\frac{4}{3}, \frac{2}{3}\right)$
Hence $\left(x_1 x_2+y_1 y_2\right)=6 \times \frac{4}{3}+0=8$