Question

The orthocentre of the triangle formed by $(8,0)$ and $(4,6)$ with the origin is:

• A. $(4,\frac{8}{3})$
• B. $(3,-4)$
• C. $(4,3)$
• D. $(3,4)$

Answer 1

The correct option is A $(4,\frac{8}{3})$

Slope of $OA$ $=$ $\frac{0}{8}$ $=$ $0$

Slope of line $\perp$ to $OA$ $=$ $-\infty$

$\therefore$ equation of line $\perp$ to $OA$ passing

through $B$ is $x=4.$

Slope of $AB$ $=$ $\frac{0-6}{8-4}$ $=$ $\frac{-6}{4}$ $=$ $\frac{-3}{2}$

Slope of $OD$ $\times$ Slope of $AB$ $=$ $-1$

Slope of $OD$ $=\frac{-1\times 2}{-3}$

$=\frac{2}{3}$

equation of line $OD$ $\Rightarrow$ $y-0=\frac{2}{3}(x-0)$

$y=\frac{2}{3}x$

$\therefore$ Intersection point of $x=4$ and $y=\frac{2}{3}x$ is

$y=\frac{2}{3}\left(4\right)=\frac{8}{3}$

$(4,\frac{8}{3})$

$\therefore$ Orthocentre of triangle is $(4,\frac{8}{3})$