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 $(1,1)$, then the equation of its third side is :

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

Answer 1

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

Let equation of side $AB:3x-2y+6=0...\left(1\right)$
and equation of side $BC:4x+5y-20=0...\left(2\right)$
$O$ is the orthocentre.
Slope of line $AB$ is $\frac{3}{2}$
Line $CE$ is perpendicular to $AB$
So, slope of line $CE$ is $-\frac{2}{3}$
$\therefore$ Equation of line $CE$ is
$(y-1)=-\frac{2}{3}(x-1)$
$\Rightarrow 2x+3y-5=0...\left(3\right)$
From Equations $\left(2\right)$ and $\left(3\right)$, we obtain ordinate C,
$x=\frac{35}{2},y=-10$

Now, slope of line $BC$ is $-\frac{4}{5}$
$\therefore$ Slope of line $AF$ is $\frac{5}{4}$
Thus, Equation of line $AF$ is
$y-1=\frac{5}{4}(x-1)$
$\Rightarrow 5x-4y-1=0...\left(4\right)$
From Equations $\left(1\right)$ and $\left(4\right)$, we obtain ordinate A,
$x=-13,y=-\frac{66}{4}$

Thus, the coordinates of $A$ is $(-13,-\frac{66}{4})$ and coordinates of $B$ is $(\frac{35}{2},-10)$
$\therefore$ Equation of $AB$ is $\frac{y+10}{x-\frac{35}{2}}=\frac{-10+\frac{66}{4}}{\frac{35}{2}+13}$
$\Rightarrow \frac{y+10}{\frac{2x-35}{2}}=\frac{\frac{-40+66}{4}}{\frac{35+26}{2}}$
$\Rightarrow \frac{2(y+10)}{2x-35}=\frac{26}{4}\times \frac{2}{61}$
$\Rightarrow 13(2x-35)=122(y+10)$
$\Rightarrow 26x-122y-1675=0$