Question

Let the incentre of $\Delta ABC$ is $Ι\left(2,5\right)$ . If $A=\left(1,13\right)$ and $B=\left(- 4,1\right)$ , then the coordinates of $C$ are

• A. $\left(1,10\right)$
• B. $\left(10,1\right)$
• C. $\left(8,2\right)$
• D. $\left(9,3\right)$

Answer 1

Answer: (B)

Slope of $AB$ is $\frac{13 - 1}{1 - \left(- 4\right)}=\frac{12}{5}$
Slope of $BI$ is $\frac{5 - 1}{2 - \left(- 4\right)}=\frac{2}{3}$
Slope of $AI$ is $\frac{13 - 5}{1 - \left(2\right)}=-8$
Let, the slopes of $BC$ and $AC$ are $m_{1}$ & $m_{2}$ respectively
$\Rightarrow \frac{m_{2} - \left(- 8\right)}{1 + m_{2} \left(- 8\right)}=\frac{- 8 - \left(\frac{12}{5}\right)}{1 + \left(- 8\right) \left(\frac{12}{5}\right)}$
$\left(m_{2} + 8\right)\left(91\right)=\left(1 - 8 m_{2}\right)\left(- 57\right)$
$8\times 91-57=\left(8 \times 57 - 91\right)m_{2}\Rightarrow m_{2}=-\frac{4}{3}$
Now, $\frac{m_{1} - \frac{2}{3}}{1 + m_{1} \left(\frac{2}{3}\right)}=\frac{\frac{2}{3} - \frac{12}{5}}{1 + \frac{2}{3} \times \frac{12}{5}}$
$\Rightarrow \left(3 m_{1} - 2\right)\left(39\right)=\left(3 + 2 m_{1}\right)\left(- 26\right)$
$\Rightarrow 9m_{1}-6=-6-4m_{1}\Rightarrow m_{1}=0$
$\Rightarrow $ The equation of $AC$ is $\left(y - 13\right)=-\frac{4}{3}\left(x - 1\right)$ and the equation of $BC$ is $y=1$
$\Rightarrow C$ is $\left(10,1\right)$