Question

The orthocentre of the triangle formed by the lines $x+3y-10=0$ and $6x^2+xy-y^2=0$ is

• A. $(1,3)$
• B. $(3,-1)$
• C. $(1,-3)$
• D. $(-1,-3)$

Answer 1

The correct option is A $(1,3)$

$6x^2+xy-y^2=0$

Substitute $t=\frac{y}{x}$

$\Rightarrow -t^2+t+6=0$

$\Rightarrow t^2-t-6=0$

$\Rightarrow t=\frac{1\pm \sqrt{1+24}}{2}$

$\Rightarrow t=\frac{1\pm 5}{2}$

$\Rightarrow t=3,-2$

And $x+3y=10$

$Slope=\frac{-1}{3}$

Two sides of triangle
i.e. $y=3x$ and $x+3y-10=0$ are perpendicular to each other

So, orthocentre of triangle is intersection of lines

$y=3x$ and $x+3y-10=0$

$\therefore x=1,y=3$

Therefore, $(1,3)$ is orthocenter.