Question

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

• A. $(-33,-26)$
• B. $(33,26)$
• C. $(26,33)$
• D. None of these

Answer 1

Answer: (B)

$(33,26)$

Let the third vertex be $(x,y)$.

Then $A=(4,-3)$, $B=(-2,5)$ , $C=(x,y)$

And Orthocentre is $I=(1,2)$.

The slope of $AB=\frac{5-(-3)}{-2-4}=\frac{-4}{3}$

Now $IC$ will be perpendicular to $AB$.

Hence

Slope of IC $=\frac{y-2}{x-1}$

Thus $\frac{y-2}{x-1}=\frac{3}{4}$

$4y-8=3x-3$

$3x-4y+5=0$ ...(i)

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

Slope of $IB$ $=\frac{5-2}{-2-1}$ $=-1$

Since $IB$ is perpendicular to $AC$ hence

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

$y+3=x-4$

$x-y-7=0$ ...(ii)

Thus we get two equations whose intersection gives the vertex C.

$x-y-7=0$

$3x-4y+5=0$

$\to x=33,y=26$

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