Question

Coordinates of the orthocentre of the triangle whose sides are $3x-2y=6,3x+4y+12=0$ and $3x-8y+12=0$ is?

Answer 1

$3x-2y=6⟶\left(i\right)$

$3x+4y=-12⟶\left(ii\right)$

$3x-8y=-12⟶\left(iii\right)$

solving $\left(i\right)\&\left(ii\right),$

$\Rightarrow 6y=-18$

$y=-3$

$x=0$

solving $\left(ii\right)\&\left(iii\right),$

$\Rightarrow 2y=0$

$y=0$

$x=-4$

solving $\left(i\right)\&\left(iii\right),$

$\Rightarrow -6y=-18$

$y=3$

$x=4$

$\therefore$ The vertices are $A(0.-3);B(-4,0);C(4,3)$

For equation of CD -

Equation of AB is given by

$\Rightarrow 3x+4y=-12$

$\Rightarrow$ Slope of $CD=\frac{4}{3}$

$\therefore$ equation of CD $\Rightarrow (y-3)=\frac{4}{3}(x-4)$

$\Rightarrow 3y-9=4x-16$

$\Rightarrow 4x-3y-7=0\to \left(iv\right)$

for equation of BE -

Equation of AC is given by

$\Rightarrow 3x-2y=6$

$\Rightarrow$ Slope of $BE=\frac{-2}{3}$

$\therefore$ equation of BE $\Rightarrow (y-0)=\frac{-2}{3}(x+4)$

$\Rightarrow 3y=-2x-8$

$\Rightarrow 2x+3y+8=0\to \left(v\right)$

Other centre is the point of intersection of $CD$ & $BE$

$\Rightarrow 4x-3y-7=0$

$2x+3y+8=0$

$6x+1=0$

$x=\frac{-1}{6}$

Putting in $\left(v\right),$

$\Rightarrow \frac{-1}{3}+3y=-8$

$\Rightarrow 3y=-8+\frac{1}{3}$

$\Rightarrow y=\frac{-23}{9}$

$\therefore$ Other centre $=(\frac{-1}{6},\frac{-23}{9})$

Hence, solved.