Question

The equation of sides $AB,BC,CA$ of a triangle are $x+y=1,4x-y+4=0$ and $2x+3y=6$ then the equation of orthocentre through the vertex $A$ is

• A. $x+4y=13$
• B. $7x-7y=13$
• C. $x-4y=13$
• D. $7x+7y=13$

Answer 1

The correct option is A $x+4y=13$

We have,

Equation of sides AB, BC, CA of a triangle are

$x+y-1=0......\left(1\right)$

$4x-y+4=0......\left(2\right)$

$2x+3y-6=0......\left(3\right)$

A line passing through intersection of (1) and (2) to

$x+y-1+\lambda (4x-y+4)=0$

$\Rightarrow (1+4\lambda )x+(1-\lambda )y+(4\lambda -1)=0......\left(4\right)$

It must be perpendicular to equation (3) to,

$ 2\left( 1+4\lambda \right)+3\left( 1-\lambda \right)=0 $

$\Rightarrow 2+8\lambda +3-3\lambda =0$

$\Rightarrow 5-5\lambda =0$

$\Rightarrow \lambda =1$

Put in (4) and we get,

$(1+4\times 1)x+(1-1)y+(4\times 1-1)=0$

$\Rightarrow 5x+0y+3=0......\left(5\right)$

Now,

A line passes through intersection of (1) and (3) to, and we get,

$(x+y-1)+\mu (2x+3y-6)=0$

$\Rightarrow (1+2\mu )x+(1+3\mu )y-1-6\mu =0......\left(6\right)$

It must be perpendicular to ( 2) and we get,

$ 4\left( 1+2\mu \right)-1\left( 1+3\mu \right)=0 $

$\Rightarrow 4+8\mu -1-3\mu =0$

$\Rightarrow 3+5\mu =0$

$\Rightarrow 5\mu =-3$

$\Rightarrow \mu =\frac{-3}{5}$

Put in ( 6) and we get,

$(1-\frac{6}{5})x+(1-\frac{9}{5})y-1+\frac{18}{5}=0$

$\Rightarrow -x-4y+13=0$

$\Rightarrow x+4y-13=0......\left(7\right)$

This is the requried equation,

So,

Solving equation (5) and (7) to, and we get,

$x=\frac{-3}{5},y=\frac{-17}{5}$

Hence, the orthocenter is $(\frac{-3}{5},\frac{-17}{5})$.

Hence, this is the answer.