If is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals [2024]

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Solution

Q.
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 + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals [2024]

1 29

2 6

3 8

4 14

Ans.
(4)

Let point P(x, y) be equidistant from the given lines.

$∴ \frac{x + 2 y + 7}{\sqrt{5}} = \pm \frac{2 x - y + 8}{\sqrt{5}}$

$\Rightarrow {(x + 2 y + 7)}^{2} = {(2 x - y + 8)}^{2}$

$\Rightarrow x^{2} + 4 y^{2} + 49 + 4 x y + 28 y + 14 x = 4 x^{2} + y^{2} + 64 - 4 x y - 16 y + 32 x$

$\Rightarrow 3 x^{2} - 3 y^{2} - 8 x y + 18 x - 44 y + 15 = 0$

$\Rightarrow x^{2} - y^{2} - \frac{8}{3} x y + 6 x - \frac{44}{3} y + 5 = 0$ ... (i)

This is the locus of the point P(x, y).

Now, compare equation (i) with given equation of locus, we get

$2 h = \frac{- 8}{3} \Rightarrow h = \frac{- 4}{3}$ ,

$2 g = 6 \Rightarrow g = 3$ ,

$2 f = \frac{- 44}{3} \Rightarrow \frac{- 22}{3}$ ,

and c = 5

$∴ g + c + h - f = 3 + 5 - \frac{4}{3} + \frac{22}{3} = 14$ .

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