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

The Correct Option is C

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 \( g + c + h - f \):

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