Question

The product of the perpendiculars from $(-1, 2)$ to the pair of lines $2x^{2} - 5xy + 2y^{2} + 3x - 3y + 1 = 0$ is

• A. $\dfrac{5}{12}$
• B. $\dfrac{12}{5}$
• C. $\dfrac{6}{5}$
• D. $\dfrac{5}{6}$

Hint:

Product of perpendicular distances from $(x_1,y_1)$ to the pair of lines $ax^2+2hxy+by^2+\cdots = 0$ equals $\dfrac{|ax_1^2+2hx_1y_1+by_1^2+2gx_1+2fy_1+c|}{\sqrt{(a-b)^2+4h^2}}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a pair of lines $ax^2+2hxy+by^2+2gx+2fy+c=0$, the product of perpendicular distances from $(x_1,y_1)$ is given by a standard formula.
Step 2: Detailed Explanation:
Here $a=2$, $h=-5/2$, $b=2$, $g=3/2$, $f=-3/2$, $c=1$.
Numerator $= |ax_1^2 + 2hx_1y_1 + by_1^2 + 2gx_1 + 2fy_1 + c|$ evaluated at $(-1,2)$.
$= |2(1) + 2(-5/2)(-1)(2) + 2(4) + 2(3/2)(-1) + 2(-3/2)(2) + 1| = |2 + 10 + 8 - 3 - 6 + 1| = 12$.
Denominator $= \sqrt{(a-b)^2 + 4h^2} = \sqrt{0 + 25} = 5$.
Product $= \dfrac{12}{5}$.
Step 3: Final Answer:
The product of perpendiculars is $\dfrac{12}{5}$.