Question

Two vertices of a triangle are $(4,-3)$ & $(-2,5)$. If the orthocentre of the triangle is at $(1,2)$, find the coordinates of the third vertex.

Answer 1

Slope$=\frac{y_2-y_1}{x_2-x_1}$

Slope of perpendicular on BC$=\frac{2-(-3)}{1-\left(4\right)}=\frac{-5}{3}$

Slope pf perpendicular on AC$=\frac{5-2}{-2-1}=-1$

$\Rightarrow$ as BC is perpendicular to OA

Slope of BC$=\frac{5-y}{-2-x}$

$\frac{(5-y)}{-(x+2)}\times \left(\frac{-5}{3}\right)=-1$

$5(y-5)=3(x+2)$

$3x-5y+31=0$ ……….(i)

$\Rightarrow$ as AC is perpendicular to OB

Slope of AC$=\frac{-3-y}{4-x}$

$\frac{-(y+3)}{(4-x)}\times (-1)=-1$

$y+3=x-4$

$x-y-7=0$ ………..(ii)

Solving (i) and (ii)

$3x-5y+31=0$

$5x-5y-35=0$

______________

$-2x+66=0$

______________

$x=33$

$33-7=y$

$\Rightarrow y=26$

$(x,y)=(33,26)$.