JEE Mains · Maths · STD 11 - 9. straight line
If \(x^2-y^2+2 h x y+2 g x+2 f y+c=0\) is the locus of a point, which moves such that it is always equidistant from the lines \(x+2 y+7=0\) and \(2 x-y\) \(+8=0\), then the value of \(\mathrm{g}+\mathrm{c}+\mathrm{h}-\mathrm{f}\) equals
- A \(14\)
- B \(6\)
- C \(8\)
- D \(29\)
Answer & Solution
Correct Answer
(A) \(14\)
Step-by-step Solution
Detailed explanation
Cocus of point \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) whose distance from Gives \(\mathrm{X}+2 \mathrm{y}+7=0 \& 2 \mathrm{x}-\mathrm{y}+8=0\) are equal is \(\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}\) \((x+2 y+7)^2-(2 x-y+8)^2=0\) Combined equation of lines…
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