Question

A triangle has its two sides along the lines $y=m_{1}x$ and $y=m_{2}x$, where $m_1,m_2$ are the roots of $2x^2-5x+2=0.$ If $(2,2)$ is the orthocentre of the triangle, then the equation of the third side is

• A. $2x-2y=9$
• B. $2x+2y=9$
• C. $x+y=9$
• D. $x-y=9$

Answer 1

The correct option is B $2x+2y=9$

$2x^2-5x+2=0$
$\Rightarrow x=\frac{1}{2},2$

Let $OAB$ be the triangle.
One vertex of the triangle is $O(0,0)$
Let equation of $OA$ be $2y=x\cdots \left(1\right)$
and equation of $OB$ be $y=2x\cdots \left(2\right)$

Let equation of $AB$ be $y=mx+c$
Orthocentre is $H(2,2)$
Slope of $OH$ is $1$
Since, $OH\perp AB$, we get $m=-1$
$y=-x+c\cdots \left(3\right)$

Solving $\left(1\right)$ and $\left(3\right),$
$A\equiv (\frac{2c}{3},\frac{c}{3})$
Now, equation of line perpendicular to $OB$ and passing through $H(2,2)$ is
$y-2=\frac{-1}{2}(x-2)$
$\Rightarrow 2y+x=6$
But this line passes through $A(\frac{2c}{3},\frac{c}{3})$
$\Rightarrow \frac{2c}{3}+\frac{2c}{3}=6$
$\Rightarrow c=\frac{9}{2}$
Hence, equation of the third side is $y=-x+\frac{9}{2}$