If x^2-y^2+2hxy+2gx+2fy+c=0 is the locus of a | vedclass

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(A) The locus of a point $P(x, y)$ equidistant from the lines $x+2y+7=0$ and $2x-y+8=0$ is given by the angle bisectors of these lines.Equating the di

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$14$

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(A) The locus of a point $P(x, y)$ equidistant from the lines $x+2y+7=0$ and $2x-y+8=0$ is given by the angle bisectors of these lines.Equating the distances:$\frac{|x+2y+7|}{\sqrt{1^2+2^2}} = \frac{|2x-y+8|}{\sqrt{2^2+(-1)^2}}$$|x+2y+7| = |2x-y+8|$$(x+2y+7)^2 - (2x-y+8)^2 = 0$$(x+2y+7 - (2x-y+8))(x+2y+7 + 2x-y+8) = 0$$(-x+3y-1)(3x+y+15) = 0$$(x-3y+1)(3x+y+15) = 0$$3x^2 + xy + 15x - 9xy - 3y^2 - 45y + 3x + y + 15 = 0$$3x^2 - 3y^2 - 8xy + 18x - 44y + 15 = 0$Dividing by $3$ to match the form $x^2-y^2+2hxy+2gx+2fy+c=0$:$x^2 - y^2 - \frac{8}{3}xy + 6x - \frac{44}{3}y + 5 = 0$Comparing coefficients:$2h = -\frac{8}{3} \implies h = -\frac{4}{3}$$2g = 6 \implies g = 3$$2f = -\frac{44}{3} \implies f = -\frac{22}{3}$$c = 5$Calculating $g+c+h-f = 3 + 5 + (-\frac{4}{3}) - (-\frac{22}{3}) = 8 - \frac{4}{3} + \frac{22}{3} = 8 + \frac{18}{3} = 8 + 6 = 14$.

If $x^2-y^2+2hxy+2gx+2fy+c=0$ 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

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