Question:medium

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

Show 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 $x$.
Updated On: Jun 8, 2026
  • $\frac{9}{5}$ units
  • $\frac{1}{5}$ units
  • $4$ units
  • $9$ units
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understand the pair of lines.
The equation $2x^2-5xy+2y^2=0$ represents two straight lines through the origin. We want the product of distances from $(2,-1)$ to these two lines.
Step 2: Factor into two lines.
Split the middle term: $2x^2-4xy-xy+2y^2=2x(x-2y)-y(x-2y)=(2x-y)(x-2y)$. So the lines are $2x-y=0$ and $x-2y=0$.
Step 3: Distance to the first line.
For $2x-y=0$ at $(2,-1)$: $\frac{|2(2)-(-1)|}{\sqrt{2^2+(-1)^2}}=\frac{|5|}{\sqrt{5}}=\frac{5}{\sqrt{5}}=\sqrt{5}$.
Step 4: Distance to the second line.
For $x-2y=0$ at $(2,-1)$: $\frac{|2-2(-1)|}{\sqrt{1^2+(-2)^2}}=\frac{|4|}{\sqrt{5}}=\frac{4}{\sqrt{5}}$.
Step 5: Multiply the two distances.
Product $=\sqrt{5}\cdot\frac{4}{\sqrt{5}}=4$.
Step 6: State the answer.
The product of the perpendicular distances is $4$ units, which is option (C).
\[ \boxed{\,4\ \text{units}\,} \]
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