Question

The coordinates of $△ABC$ are $A(0,2),B(-2,6)$ and $C(4,0)$. Find the coordinates of the orthocentre of this triangle.

• A. $(1,2)$
• B. $(8,-2)$
• C. $(-1,-2)$
• D. $(12,10)$

Answer 1

The correct option is D $(12,10)$

Slope of $AC=\frac{0-2}{4-0}=\frac{-1}{2}$

$BD\perp AC$

Let the slope of $BD$ be $m$

We know, product of the slopes of the perpendicular lines is $-1$

$\Rightarrow$ $m\times \frac{-1}{2}=-1$

$\Rightarrow$ $m=2$

Equation of $BD=(y-y_1)=m(x-x_1)$

$\Rightarrow$ $(y+2)=2(x-6)$

$\Rightarrow$ $y+2=2x-12$

$\Rightarrow$ $2x-y=14$ ----- ( 1 )

Slope of $AB=\frac{6-2}{-2-0}=-2$

Let the slope of $CE$ be $n$

$\Rightarrow$ $n\times (-2)=-1$

$\Rightarrow$ $n=\frac{1}{2}$

Equation of $CE=(y-y_1)=n(x-x_1)$

$\Rightarrow$ $(y-4)=\frac{1}{2}(x-0)$

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

$\Rightarrow$ $x-2y=-8$ ------ ( 2 )

Point of intersection of the perpendicular is the ortho-center of the triangle.

Solving equation of ( 1 ) and ( 2 ) we get,

$x=12$ and $y=10$

$\therefore$ Ortho-center is $(12,10)$