Question

If the product of the perpendicular distance from $(1,k)$ to the pair of lines $x^2-4xy+y^2=0$ is $\frac{3}{2}$ units, then $k=$

• A. $4$
• B. $5$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$5$

The equation of pair of lines $x^2-4xy+y^2$ is written in terms of slope of the pair of lines as $m^2-4m+1=0$.

Solving the quadratic equation we obtain the slopes of each line as $m_1=2+\sqrt{3}$ and $m_2=2-\sqrt{3}$.

Hence we obtain the equation of each of the straight lines as

$y=(2+\sqrt{3})x$, $y=(2-\sqrt{3})x\left(a\right)$

Formula for distance from a point $(x_1,y_1)$ to a line $ax+by+c=0$ is given as $\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

The product of distances from point $(1,k)$ to the pair of lines given in $\left(a\right)$ is given as

$\frac{|-(2+\sqrt{3})+k|}{\sqrt{(2+\sqrt{3}{)}^2+1}}.\frac{|-(2-\sqrt{3})+k|}{\sqrt{(2-\sqrt{3}{)}^2+1}}=\frac{3}{2}$

By expanding the above expression we obtain $\frac{|1-4k+k^2|}{\sqrt{16}}=\frac{3}{2}$

By solving the quadratic equation in terms of k, we obtain $|(k+1)(k-5)|=0$. We have solution $k=5$ in option B.