Question

If $x^2-y^2+2hxy+2gx+2fy+c=0$ is the locus of points such that it is equidistant from the lines $x+2y-8=0$ and $2x+y+7=0$, then value of $g+c+h-f$ is

• A. 29
• B. 6
• C. 14
• D. 8

Answer 1

Answer: (C)

14

Answer 2

The locus of the point $P(x, y)$, whose distance from the lines $x + 2y + 7 = 0$ and $2x - y + 8 = 0$ is equal, is given by the equation:

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

Simplifying, we get:

$(x + 2y + 7)^2 = (2x - y + 8)^2.$

For the combined equation of lines, we have:

$(x - 3y + 1)(3x + y + 15) = 0.$

Expanding, we get:

$3x^2 - 3y^2 - 8xy + 18x - 44y + 15 = 0.$

Rewriting in standard form:

$x^2 - y^2 - \frac{8}{3}xy + 6x - \frac{44}{3}y + 5 = 0.$

Thus, the equation becomes:

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

where we identify:

$h = \frac{4}{3}, \quad g = 3, \quad f = -\frac{22}{3}, \quad c = 5.$

Now, calculate $g + c + h - f$:

$g + c + h - f = 3 + 5 + \frac{4}{3} + \frac{22}{3} = 8 + 6 = 14.$