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Q.
The orthocentre of the triangle whose vertices are $(5,-2)$, $(-1,2)$ and $(1,4)$ is
Straight Lines
Solution:
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 $\bot$ line $AD = - 1$
$\therefore $ Equation of line $AD$ passing through $\left(5, - 2\right)$ is
$y + 2 = - 1\left(x - 5\right)$
$\Rightarrow x + y - 3 = 0\quad\ldots\left(i\right)$
Slope of line $AC=\left(\frac{4+2}{1-5}\right)=\left(\frac{6}{-4}\right)=-\frac{3}{2}$
Slope of $\bot$ 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 $2x - 3y = - 8 \quad\ldots\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$,
we get $x =\frac{1}{5}$,
$y=\frac{14}{5}$
$\therefore $ Orthocentre is $\left(\frac{1}{5}, \frac{14}{5}\right)$.
