Question

The equation of two sides of a $△$ are $3x-2y+6=04x+5y-20=0$ resp. The orthocentre of $△$ is $(1,1)$ find equation of third side which lies on the point $(\frac{35}{2},-10)$ ?

Answer 1

Let equation of AB be : $3x-2y+6=0$ ___(1)

Let equation of AC be : $4x+5y-20=0$ __(ii)

$3x-2y+6=0\Rightarrow x=\frac{2y-6}{3}$ ___ from (i)

replacing value of $x$ in (ii)

$4\times \frac{(2y-6)}{3}+5y-20=0$

$\Rightarrow 8y-24+15y-60=0$

$\Rightarrow 23y=84\Rightarrow y=\frac{84}{23}$

$\therefore (x,y)=(\frac{10}{23},\frac{89}{23})$

now, $x=\frac{2\times \frac{89}{23}-6}{3}=\frac{168-138}{3\times 23}=\frac{30}{3\times 23}=\frac{10}{23}$

$\therefore$ equation of alt. AP $\Rightarrow \frac{y-1}{x-1}=\frac{\frac{84}{23}-1}{\frac{10}{23}-1}$

$\Rightarrow \frac{y-1}{x-1}=\frac{61}{-13}\Rightarrow -13y+13=61x-61$

$\Rightarrow 61x+13y-74=0$

$\because BC\perp AP$

$\therefore m\left(BC\right)=\frac{-1}{m_{\left(AP\right)}}$

now, for $m_{AP}$

equation of AP $\Rightarrow 61x+13y-74=0$

$\Rightarrow 13y=74-61x\Rightarrow y=\frac{-61}{13}x+\frac{74}{13}$

$\therefore m_{BC}=\frac{-1}{-\frac{61}{13}}=\frac{13}{61}$

$\therefore$ equation of $BC\Rightarrow y=\frac{13}{61}x+c$

$[\because y=mx+c]$

$\Rightarrow 61y-13x=c$

As, BC lies on $(\frac{35}{2},10)$

$\therefore$ equation of $BC\Rightarrow 61\cdot (-10)-13\cdot \frac{35}{2}=c=\frac{-1675}{2}$

$\therefore$ equation of BC becomes $\Rightarrow 61y-13x+\frac{1675}{2}=0$

$\Rightarrow 13x-61y-\frac{1675}{2}=0$