Question

Two vertices of a triangle are B(4, -3) C(-2, 5). If the orthocenter of the triangle is (1 , 2) find the co- ordinates of the third vertex.

Answer 1

A lies on the line perpendicular to BC passing through ortho center

Slope of $BC=\frac{8}{-6}=-\frac{4}{3}$

$\therefore$Slope of required line$=\frac{3}{4}$

Equation of required line$=(y-2)=\frac{3}{4}(x-1)$

$=4y-8=3x-3$

$=3x-4y+5=0$

Let $A(x_1,y_1)$ be the third co ordinate

$\therefore 3x_1=-5+4y_1$

$\Rightarrow x_1=\frac{4y_1-5}{3}$

$\therefore A=(\frac{4y_1-5}{3},y_1)$

Now, if O is the orthocenter, $BO\perp AC$

$\Rightarrow \frac{-5}{3}\times \frac{y_1-5}{\frac{4y_1-5}{3}+2}=-1$

$\Rightarrow 5y1-25=4y_1+1\Rightarrow y_1=26$

$\therefore x_1=\frac{99}{3}=33$

$\therefore$ Third vertex is $(33,26)$

Hence, this is the answer.