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: Understanding the Question:
We need to find the product of distances from a point $(x_1, y_1)$ to the two lines represented by the joint equation $ax^2 + 2hxy + by^2 = 0$.

Step 2: Key Formula or Approach:
Factor the equation $2x^2 - 5xy + 2y^2 = 0$ into two separate linear equations $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$. Calculate the distance from $(2, -1)$ to each and multiply.

Step 3: Detailed Explanation:
Factorize: $2x^2 - 4xy - xy + 2y^2 = 2x(x - 2y) - y(x - 2y) = (2x - y)(x - 2y) = 0$.
The lines are $L_1: 2x - y = 0$ and $L_2: x - 2y = 0$.
Distance $d_1$ from $(2, -1)$ to $L_1$: $\frac{|2(2) - (-1)|}{\sqrt{2^2 + (-1)^2}} = \frac{|4 + 1|}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$.
Distance $d_2$ from $(2, -1)$ to $L_2$: $\frac{|1(2) - 2(-1)|}{\sqrt{1^2 + (-2)^2}} = \frac{|2 + 2|}{\sqrt{5}} = \frac{4}{\sqrt{5}}$.
Product of distances = $d_1 \cdot d_2 = \sqrt{5} \cdot \frac{4}{\sqrt{5}} = 4$.

Step 4: Final Answer:
The product of the distances is 4, which is option (C).