Question

The orthocenter of the triangle formed by the pair of lines $2x^2-xy-y^2+x+2y-1=0$ and the line $x+y+1=0$

• A. (-1,0)
• B. (0, 1)
• C. (1, 1)
• D. None of these

Answer 1

Answer: (A)

(-1,0)

Given pair of lines are$2x^2-xy-y^2+x+2y-1=0$

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

$(x-y)(2x+y)+x+2y-1=0$

Equations are $x-y+c_1=0and2x+y+c_2=0$

Also

$2x^2-xy-y^2+x+2y-1=0=(x-y+c_1)(2x+y+c_2)$

=$2x^2+yx+x{c}_2-2xy-y^2-y{c}_2+2x{c}_1+y{c}_1+c_1{c}_2$

=$2x^2-xy-y^2+x(c_2+2c_1)+y(c_1-c_2)+c_1{c}_2$

On comparing we get

$c_2+2c_1=1$ and $c_1-c_2=2$

On solving we get $c_1=1and{c}_2=-1$

Eqs.(1)$\to x-y+1=0-\left(2\right)$

and $2x+y-1=0-\left(3\right)$

Here slope of line (1) and slope of line (2) are -1 and 1

Product of slope =-1

Hence line 1$\perp$line 2

So the intersection of line 1 and line 2 forming a right angle which is an orthocenter

solving eqs.(1) and (2) we get, x=-1 and y=0

So orthocenter is (-1,0)