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

Solution

Maths sir · 4 steps

Equidistant from two lines means the locus is their pair of angle bisectors: square once and compare.

1. Equal distances. Both lines have √(1²+2²) = √(2²+1²) = √5 in the denominator of the distance formula, so it cancels. The condition becomes |x+2y+7| = |2x-y+8|. Squaring removes the modulus and covers both bisectors at once.

   (x+2y+7)² = (2x-y+8)²

2. Expand both sides. The left side is x²+4y²+4xy+14x+28y+49. The right side is 4x²+y²-4xy+32x-16y+64. Moving everything to the right gives 3x²-3y²-8xy+18x-44y+15=0.
3. Match the form. Divide by 3 so the x² coefficient is 1: x²-y²-8/3xy+6x-44/3y+5=0. Comparing, 2h=-8/3, 2g=6, 2f=-44/3 and c=5. So h=-4/3, g=3, f=-22/3 and c=5.
4. So. g+c+h-f = 3+5-4/3+22/3 = 8+6 = 14, which is option (b).

The idea to keep. A point equidistant from two lines lies on their angle bisectors. Squaring the distance equation gives both bisectors as one quadratic. Scale it to the given form before you compare coefficients.

Answer from the NTA final answer key. Solution by PrepWiser.