Question

The equation of two sides of triangle are $3x-2y+6=0$ and $4x+5y=20$ and orthocentre is $(1,1)$, then the equation of third side is

• A. $26x-122y-1675=0$
• B. $13x-61y-837=0$
• C. $26x+122y+1675=0$
• D. $13x+61y+837=0$

Answer 1

The correct option is A $26x-122y-1675=0$


Let $AB:3x-2y+6=0,$ $AC:4x+5y=20$

Let $A(x,y)$ be the point of intersection of $AB$ and $AC$

$\Rightarrow x=\frac{10}{23},y=\frac{84}{23}$

Given $O(1,1)$ is orthocentre of $△ABC.$

Let $AD$ and $BE$ be the altitudes from $A$ and $B$ respectively

$\Rightarrow AD\perp BC$ and $BE\perp AC$

$\Rightarrow m_{AD}\times m_{BC}=-1$

$\Rightarrow \frac{\frac{84}{23}-1}{\frac{10}{13}-1}\times m_{BC}=-1$

$\Rightarrow m_{BC}=\frac{13}{61}$

$m_{BC}=\frac{13}{61}$

Now, $BE\perp AC$ (where slope of $AC=-\frac{4}{5})$ $\Rightarrow m_{BE}\times m_{AC}=-1$
Let $B=(\alpha ,\beta )$

$\Rightarrow \frac{\beta -1}{\alpha -1}\times -\frac{4}{5}=-1$

$\Rightarrow 5\alpha -4\beta =1$

$m_{BC}=\frac{13}{61}$ and $5\alpha -4\beta =1$

Also, $B$ lies on $AB:3x-2y+6=0$

$\Rightarrow 3\alpha -2\beta +6=0$

$5\alpha -4\beta -1=0$

$6\alpha -4\beta +12=0$

$\alpha =-13$

$\beta =-\frac{33}{2}$

By point slope form equation of $BC$ is
$\frac{y+\frac{33}{2}}{x+13}=\frac{13}{61}$
$\Rightarrow 26x-122y-1675=0$