If the solve the problem


If a $\triangle \mathrm{ABC}$ has vertices $\mathrm{A}(-1,7), \mathrm{B}(-7,1)$ and $\mathrm{C}(5,-5)$, then its orthocentre has coordinates:

  1. $(3,-3)$

  2. $\left(-\frac{3}{5}, \frac{3}{5}\right)$

  3. $(-3,3)$

  4. $\left(\frac{3}{5},-\frac{3}{5}\right)$

Correct Option: , 3


Let orthocentre is $\mathrm{H}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$

$\mathrm{m}_{\mathrm{AH}} \cdot \mathrm{m}_{\mathrm{BC}}=-1$


$\Rightarrow \quad 2 \mathrm{x}_{0}-\mathrm{y}_{0}+9=0$ ......(1)

and $\mathrm{m}_{\mathrm{BH}} \cdot \mathrm{m}_{\mathrm{AC}}=-1$

$\Rightarrow \quad\left(\frac{\mathrm{y}_{0}-1}{\mathrm{x}_{0}+7}\right)\left(\frac{7-(-5)}{-1-5}\right)=-1$

$\Rightarrow \quad x_{0}-2 y_{0}+9=0$ .....(2)

Solving equation (1) and (2) we get

$\left(x_{0}, y_{0}\right) \equiv(-3,3)$

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