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Q.
Orthocentre of triangle with vertices $(0, 0), (3, 4)$ and $(4, 0)$ is
Straight Lines
Solution:
To find orthocentre of the triangle formed by $(0, 0) (3, 4) $ and $(4, 0)$.
Let $H$ be the orthocentre of $\Delta OAB$
$\therefore $ (slope of $OP$ i.e. $OH$) -(slope of $BA) = -1$
$\Rightarrow \Big(\frac{y-0}{3-0}\Big).\Big(\frac{4-0}{3-4}\Big)=-1$
$\Rightarrow -\frac{4}{3}y=-1$
$\Rightarrow y=\frac{4}{3}$
$\therefore $ Required orthocentre $= (3, y) =\Big(3,\frac{3}{4}\Big)$
