Question

Let the equations of two sides of a triangle be $3x-2y+6=0$ and $4x+5y-20=0$. If the orthocentre of this triangle is at $(1,1)$ and the equation of its third side is $ax+by+c=0$, then $a+b-c$ is

Answer 1

Answer: 1579

Since, $AH$ is perpendicular to $BC$
Hence, $m_{AH}.m_{BC}=-1$
$\left(\frac{\frac{20-4x_2}{5}-1}{x_2-1}\right)\times \frac{3}{2}=-1$
$\frac{15-4x_2}{5\left(x_2-1\right)}=-\frac{2}{3}$
$45-12x_2=-10x_2+10$
$2x_2=35\Rightarrow x_2=\frac{35}{2}$
$\Rightarrow A\left(\frac{35}{2},-10\right)$
Since, $BH$ is perpendicular to $CA$.
Hence, $m_{BH}\times m_{CA}=-1$
$\left(\frac{\frac{3x_1}{2}+3-1}{x_1-1}\right)\left(-\frac{4}{5}\right)=-1$
$\frac{\left(3x_1+4\right)}{2\left(x_1-1\right)}\times 4=5$
$\Rightarrow 6x_1+8=5x_1-5$
$\Rightarrow x_1=-13$
$\Rightarrow \left(-13,\frac{-33}{2}\right)$
$\Rightarrow$ Equation of line $AB$ is
$y+10=\left(\frac{\frac{-33}{2}+10}{-13-35}\right)\left(x-\frac{35}{2}\right)$
$\Rightarrow -16y-610=-13x+\frac{455}{2}$
$\Rightarrow -122y-1220=-26x+455$
$\Rightarrow 26x-122y-1675=0$
Equation of third sideis $26x-122y-1675=0$
$\Rightarrow ax+by+c=26x-122y-1675$
$a=26,b=-122,c=-1675$
$\therefore a+b-c=26+(-122)-(-1675)$
$=1579$