Question

The product of the perpendicular distances from $(2,-1)$ to the pair of lines $2 x^2-5 x y+2 y^2=0$ is

• A. $\frac{9}{5}$ units
• B. $\frac{1}{5}$ units
• C. $4$ units
• D. $9$ units

Hint:

For a joint equation $ax^2 + 2hxy + by^2 = 0$, you can find the individual lines by solving the quadratic for $y$ in terms of $x$.

Answer 1

The Correct Option is C

Step 1: Split the pair into two lines.
Factor $2x^2-5xy+2y^2 = (2x-y)(x-2y)$. So the two lines are $2x-y=0$ and $x-2y=0$.

Step 2: Distance to the first line.
From $(2,-1)$ to $2x-y=0$: \[ \frac{|2(2)-(-1)|}{\sqrt{4+1}} = \frac{5}{\sqrt5} = \sqrt5. \]

Step 3: Distance to the second line.
From $(2,-1)$ to $x-2y=0$: \[ \frac{|2-2(-1)|}{\sqrt{1+4}} = \frac{4}{\sqrt5}. \]

Step 4: Multiply.
$\sqrt5 \times \tfrac{4}{\sqrt5} = 4$. \[ \boxed{4\ \text{units}} \]