Question

If a $△ABC$ has vertices $A(-1,7),B(-7,1)$ and $C(5,-5)$, then its orthocentre has coordinates:

• A. $(-3,3)$
• B. $(-\frac{3}{5},\frac{3}{5})$
• C. $(\frac{3}{5},-\frac{3}{5})$
• D. $(3,-3)$

Answer 1

The correct option is A $(-3,3)$

Let coordinates of orthocentre be $H(x,y)$

$m_{BC}=\frac{6}{-12}=-\frac{1}{2}$
$\therefore$ Equation of $AD$ is
$\Rightarrow y-7=2(x+1)$
$\Rightarrow 2x-y+9=0\cdots \left(1\right)$

$m_{AC}=\frac{12}{-6}=-2$
$\therefore$ Equation of $BE$ is
$y-1=\frac{1}{2}(x+7)$
$\Rightarrow x-2y+9=0\cdots \left(2\right)$
Solving equation $\left(1\right)$ and $\left(2\right)$, we get
$(x,y)\equiv (-3,3)$