Question

The orthocentre of the triangle whose vertices are $(5,-2)$, $(-1,2)$ and $(1,4)$ is

• A. $(1/5,14/5)$
• B. $(14/5,1/5)$
• C. $(1/5,1/5)$
• D. $(14/5,14/5)$

Answer 1

Answer: (A)

Let $P$ be the point of intersection of altitudes $AD$ and $BE$.



Since orthocentre is $P(x,y)$.
Slope of line $BC=\frac{4-2}{1-\left(-1\right)}=1$
$\therefore$ Slope of $\perp$ line $AD=-1$
$\therefore$ Equation of line $AD$ passing through $\left(5,-2\right)$ is
$y+2=-1\left(x-5\right)$
$\Rightarrow x+y-3=0\dots \left(i\right)$
Slope of line $AC=\left(\frac{4+2}{1-5}\right)=\left(\frac{6}{-4}\right)=-\frac{3}{2}$
Slope of $\perp$ line $BE=\frac{2}{3}$
$\therefore$ Equation of line $BE$ passing through $\left(-1,2\right)$ is
$y-2=\frac{2}{3}\left(x+1\right)$ or $3y-6=2x+2$
or $2x-3y=-8\dots \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$,
we get $x=\frac{1}{5}$,
$y=\frac{14}{5}$
$\therefore$ Orthocentre is $\left(\frac{1}{5},\frac{14}{5}\right)$.