Question

Write the coordinates of the orthocentre of the triangle formed by the lines $x^2-y^2=0$ and $x+6y=18$.

Answer 1

Sides of triangle are $x^2-y^2=0$

$\Rightarrow (x-y)=0,x+y=0$ and $x+6y=18$

Slope of AB=1

Slope of CN=-1

Point C is obtained by solving the equations of AC and BC

Coordinates of C is $(-\frac{18}{5},\frac{18}{5})$

Equation of CN is

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

$\frac{5y-18}{5}=\frac{-5x-18}{5}$

$5y-18=-5x-18$

$x+y=0$ ...(1)

Now, slope of line AC=-1

Slope of line BM=1

Coordinates of B is obtained by solving equations of AB and BC

$\Rightarrow x=\frac{18}{7},y=\frac{18}{7}$

Coordinates of B is$(\frac{18}{7},\frac{18}{7})$

Equation of BM is

$y-\frac{18}{7}=1(x-\frac{18}{7})$

$y-\frac{18}{7}=x-\frac{18}{7}$

$x-y-0$ ...(2)

Solving equation (1) and (2) we get Orthoecentre=(0,0)