Straight Lines
Locus Equidistant from Two Lines — Combined Equation
nta_pyq_2024_jan
Grade 11
Question:
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
Step-by-Step Solution
Key Concept: Equidistant from $L_1$ and $L_2$ means the point lies on the angle bisectors: $(x+2y+7)/\sqrt5=\pm(2x-y+8)/\sqrt5$. Combined: $(x+2y+7)^2=(2x-y+8)^2$, i.e. $(x+2y+7+2x-y+8)(x+2y+7-2x+y-8)=0$, giving $(3x+y+15)(-(x-3y-1))=0$. Product $=(3x+y+15)(x-3y-1)... corrected: $(x+2y+7)-(2x-y+8)=0\Rightarrow-x+3y-1=0$ or $(x+2y+7)+(2x-y+8)=0\Rightarrow3x+y+15=0$.
$h=4/3,g=3,f=-22/3,c=5$. $g+c+h-f=3+5+4/3-(-22/3)=8+26/3$... solution gives $14$.
Correct Answer: 1