Prove the following
Question:

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

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

  2. (2) $(-3,3)$

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

  4. (4) $(3,-3)$


Correct Option: , 2

Solution:

$m_{B C}=\frac{6}{-12}=-\frac{1}{2}$

$\therefore$ Equation of AS is $y-7=2(x+1)$

$y=2 x+9$ ….(i)

$m_{A C}=\frac{12}{-6}=-2$

$\therefore$ Equation of $B P$ is $y-1=\frac{1}{2}(x+7)$

$y=\frac{x}{2}+\frac{9}{2}$ …(ii)

From equs. (i) and (ii),

$2 x+9=\frac{x+9}{2}$

$\Rightarrow 4 x+18=x+9$

$\Rightarrow 3 x=9 \Rightarrow x=-3$

$\therefore y=3$

 

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