Question

If x²- y² + 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

• A. 29
• B. 8
• C. 14
• D. 6

Answer 1

Answer: (C)

14

Answer 2

The locus of a point equidistant from two lines is the pair of angle bisectors. The equation of the locus is formed by equating the distances from a point $(x, y)$ to the two lines.

The distance from $(x, y)$ to $x + 2y + 7 = 0$ is $\frac{|x + 2y + 7|}{\sqrt{1^2 + 2^2}} = \frac{|x + 2y + 7|}{\sqrt{5}}$. The distance from $(x, y)$ to $2x - y + 8 = 0$ is $\frac{|2x - y + 8|}{\sqrt{2^2 + (-1)^2}} = \frac{|2x - y + 8|}{\sqrt{5}}$.

Equating these distances: $\frac{|x + 2y + 7|}{\sqrt{5}} = \frac{|2x - y + 8|}{\sqrt{5}}$ $|x + 2y + 7| = |2x - y + 8|$ This gives two possibilities:

1. $x + 2y + 7 = 2x - y + 8 \implies x - 3y + 1 = 0$
2. $x + 2y + 7 = -(2x - y + 8) \implies 3x + y + 15 = 0$

The locus is the product of these two lines: $(x - 3y + 1)(3x + y + 15) = 0$ Expanding this: $3x^2 + xy + 15x - 9xy - 3y^2 - 45y + 3x + y + 15 = 0$ $3x^2 - 3y^2 - 8xy + 18x - 44y + 15 = 0$ To match the given equation $x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$, we divide by 3: $x^2 - y^2 - \frac{8}{3}xy + 6x - \frac{44}{3}y + 5 = 0$ Comparing coefficients: $2h = -\frac{8}{3} \implies h = -\frac{4}{3}$ $2g = 6 \implies g = 3$ $2f = -\frac{44}{3} \implies f = -\frac{22}{3}$ $c = 5$

Now, calculate $g + c + h - f$: $g + c + h - f = 3 + 5 + \left(-\frac{4}{3}\right) - \left(-\frac{22}{3}\right) = 8 - \frac{4}{3} + \frac{22}{3} = 8 + \frac{18}{3} = 8 + 6 = 14$