From the point (3, -4) perpendicular lines L1 and L2 are drawn on each of the lines S≡2x2+3xy-2y2-7x+y+3=0. The area of the quadrilateral formed by the pair of lines S=0, L1 and L2 is (in square units)

TS EAMCET · Maths · Pair of Lines

From the point $(3, - 4)$ perpendicular lines $L_{1}$ and $L_{2}$ are drawn on each of the lines $S \equiv 2 x^{2} + 3 x y - 2 y^{2} - 7 x + y + 3 = 0 .$ The area of the quadrilateral formed by the pair of lines $S = 0, L_{1}$ and $L_{2}$ is (in square units)

A $\frac{64}{5}$
B $\frac{72}{5}$
C $25$
D $35$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{72}{5}$

Step-by-step Solution

Detailed explanation

S≡2x2+3xy-2y2-7x+y+3=0 Consider, 2x2+3xy-2y2=0 ⇒x+2y2x-y=0 ⇒x+2y=0, 2x-y=0 Now, x+2y+c12x-y+c2≡S Comparing coefficient of x & y 2c1+c2=-7, 2c2-c1=1 ⇒c2=-1, c1=-3 Equation of lines perpendicular to S=0 & passing through 3,-4 are 2x-y+k1=0, x+2y+k2=0…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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