Question

The triangle formed by $x+3y=1$ and $9x^2-12xy+k{y}^2=0$ is right angled triangle and $k\ne -9,$ then $k=$

• A. $3$
• B. $5$
• C. $7$
• D. $1$

Answer 1

Answer: (A)

$3$

$x+3y=1$

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

$9x^2-12xy+k{y}^2=0----\left(1\right)$

Let $y=m_{1}x$ and $y=m_{2}x$ are two straight lines of

$(y-m_{1}x)(y-m_{2}x)=0$

$y^2-(m_1+m_2)xy+m_1{m}_2{x}^2=0$ ---(2)

As given $x+3y=1$ formed right angle with one of the line

$9x^2-12xy+k{y}^2=0$

So, let,
$m_1=3$

So, compare 1) & 2)

$m_1+m_2=\frac{12}{k}\Rightarrow m_2+3=\frac{12}{k}$

$m_1{m}_2=\frac{9}{k}\Rightarrow m_2=\frac{3}{k}$

$\Rightarrow \frac{3}{k}+3=\frac{12}{k}$

$\Rightarrow 3=\frac{9}{k}$

$\Rightarrow k=3$