TS EAMCET · Maths · Pair of Lines
Let \(d\) be the distance between the parallel lines \(3 x-2 y+5=0\) and \(3 x-2 y+5+2 \sqrt{13}=0\). Let \(\mathrm{L}_1 \equiv 3 \mathrm{x}-2 \mathrm{y}+\mathrm{k}_1=0\left(\mathrm{k}_1>0\right)\) and \(\mathrm{L}_2 \equiv 3 \mathrm{x}-2 \mathrm{y}\) \(+\mathrm{k}_2=0\left(\mathrm{k}_2>0\right)\) be two lines that are at the distance of \(\frac{4 \mathrm{~d}}{\sqrt{13}}\) and \(\frac{3 \mathrm{~d}}{\sqrt{13}}\) from the line \(3 \mathrm{x}-2 \mathrm{y}+5=0\). Then the combined equation of the lines \(\mathrm{L}_1=0\) and \(\mathrm{L}_2=0\) is
- A \((3 x-2 y)^2+24(3 x-2 y)+143=0\)
- B \((3 x-2 y)^2+8(3 x-2 y)+33=0\)
- C \((3 x-2 y)^2+12(3 x-2 y)+13=0\)
- D \((3 x-2 y)^2+12(3 x-2 y)+1=0\)
Answer & Solution
Correct Answer
(A) \((3 x-2 y)^2+24(3 x-2 y)+143=0\)
Step-by-step Solution
Detailed explanation
Distance between \(3 x-2 y+5=0\) and \[ \begin{aligned} & 3 x-2 y+5+2 \sqrt{13}=0 \\ & d=\frac{5+2 \sqrt{13}-5}{\sqrt{3^2+2^2}}=\frac{2 \sqrt{13}}{\sqrt{13}}=2 \end{aligned} \] \(\mathrm{L}_1: 3 x-2 y+\mathrm{k}_1 ; \mathrm{L}_2: 3 x-2 y+\mathrm{k}_2\) L: \(3 x-2 y+5\) Distance…
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