JEE Main 2020 — Straight Lines Question with Solution

Slope of line $BC$ is $m_{BC}=\frac{6}{-12}=-\frac{1}{2}$

Slope of line $AD=\frac{1}{m_{BC}}=2$

$\therefore$ Equation of line $AD$ is $y-7=2\left(x+1\right)$

$\Rightarrow y=2x+9 ...\left(1\right)$

Slope of line $AC$ is $m_{AC}=\frac{12}{-6}=-2$

Slope of line $BE=\frac{1}{m_{AC}}=\frac{1}{2}$

$\therefore$ Equation of $BE$ is $y-1=\frac{1}{2}\left(x+7\right)$

$\Rightarrow y=\frac{x}{2}+\frac{9}{2} ...\left(2\right)$

Comparing equations $\left(1\right) \& \left(2\right)$, we get

$2x+9=\frac{x+9}{2}$

$\Rightarrow 4x+18=x+9$

$\Rightarrow 3x=-9\Rightarrow x=-3$

$\therefore y=2x+9=3$

Therefore, orthocentre has co-ordinates $\left(-3,3\right)$.