TS EAMCET · Maths · Pair of Lines

If $L$ is a line passing through the point $(- 1, 1)$ and parallel to the common line of the pairs of lines $6 x^{2} - x y - 12 y^{2} = 0$ and $15 x^{2} + 14 x y - 8 y^{2} = 0$ , then the equation of pair of lines joining the origin to the points of intersection of the curve $2 x^{2} - x y - y^{2} + x - y = 0$ and the line $L$ is

A $x^{2} - x y - y^{2} = 0$
B $x^{2} + x y - y^{2} = 0$
C $x^{2} - y^{2} = 0$
D $2 x^{2} + 3 x y - 6 y^{2} = 0$

Verified Solution

Answer & Solution

Correct Answer

(C) $x^{2} - y^{2} = 0$

Step-by-step Solution

Detailed explanation

Given, Line 6x2-xy-12y2=0 ⇒6x2-9xy+8xy-12y2=0 ⇒3x+4y2x-3y=0 ..........1 Now solving 15x2+14xy-8y2=0 we get, ⇒15x2+20xy-6xy-8y2=0 ⇒5x-2y3x+4y=0 ......2 From equation 1 & 2 we get common line as 3x+4y=0 Now line passing through the…

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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If L is a line passing through the point -1,1 and parallel to the common line of the pairs of lines 6x2-xy-12y2=0 and 15x2+14xy-8y2=0, then the equation of pair of lines joining the origin to the points of intersection of the curve 2x2-xy-y2+x-y=0 and the line L is

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