Mathematics - Point Question with Solution | TestHub

Given the equations of the sides $AB,BC$ and $CA$ of a triangle $ABC$ are $2x+y=0,x+py=15a$ and $x-y=3$ respectively.

Now on solving equation $2x+y=0,x+py=15a$ and $x-y=3$ we get coordinates of $A\left(1,-2\right),B\left(\frac{15a}{1-2p},\frac{-30a}{1-2p}\right)$ and $C=\left(\frac{18p-30}{p+1},\frac{15p-33}{p+1}\right)$

And let orthocentre be $H\left(2,a\right)$

Slope of $AH=\frac{a+2}{1}$

Sloe of $BC=-\frac{1}{p}$

As $AH\perp BC$, so $p=a+2$

Now coordinate of $C=\left(\frac{18p-30}{p+1},\frac{15p-33}{p+1}\right)$

So, slope of $HC$

$=\frac{\frac{15p-33}{p+1}-a}{\frac{18p-30}{p+1}-2}=\frac{15p-33-\left(p-2\right)\left(p+1\right)}{18p-30-2p-2}$

$=\frac{16p-p^2-31}{16p-32}$

Now $HC\perp AB$, so $\text{slope of} HC\times \text{slope of }AB\text{=-1}$

$\Rightarrow \frac{16p-p^2-31}{16p-32}\times -2=-1$

$\Rightarrow p^2-8p+15=0$

$\Rightarrow p=3 \mathrm{or} 5$

But if $p=5$ then $a=3$ not acceptable as $p=a+2$

$\therefore p=3$