Question

Two vertices of a triangle are $(5,-1)$ and $(-2,3)$ if the orthocentre of the triangle is $(0,0)$ then thired vertex is-

• A. $(3/2,1)$
• B. $(1,2/3)$
• C. $(-4,-7)$
• D. $(-3,-2)$

Answer 1

The correct option is C $(-4,-7)$

$A(5,-1),B(-2,3)$

orthocentre $O(0,0)$

Slope of altitude coming from $A=\frac{-1-0}{5-0}=\frac{-1}{5}$

$\therefore$ Slope of $BC=-\left[\frac{1}{\frac{-1}{5}}\right]=5$

$\therefore$ Slope of $BC$ & slope pf altitude coming from $A=-1$

enq. of line thro $BC$ is

$y=5x+c$

It $B(-2,3)$ lies on above line

$3=-5\times 2+C$

$C=13$

$y=5x+13$ ....(1)

Slope of altitude coming from $B=\frac{3-0}{-2-0}=\frac{-3}{2}$

$\therefore$ Slope of $AC=\frac{2}{3}$

eqn. of $AC$ is $y=\frac{2}{3}x+C$

subtracts $(5,-1)$, we get $-1-\frac{10}{3}=C$

$C=\frac{-13}{3}$

$y=\frac{2}{3}x-\frac{13}{3}$ ...(2)

Eqn. (1) - (2)

$0=[5x-\frac{2x}{3}]+13-\left(\frac{-13}{3}\right)$

$0=\frac{13x}{3}+\frac{37+13}{3}$

$\frac{13}{3}x+\frac{52}{3}=0$

$13x=-52$

$x=-4$

Put value of $x$ in eqn. (1)

$y=5(-4)+13$

$y=-2a+13$

$y=-7$

$\therefore$ third vertex is $(-4,-7)$