Question:
<p>If <span class="math-tex">\(x^{2}-y^{2}+2 h x y+2 g x+2 f y+c=0\)</span> is the locus of a point, which moves such that it is always equidistant from the lines <span class="math-tex">\(x+2 y+7=\)</span> 0 and <span class="math-tex">\(2 x-y+8=0\)</span>, then the value of <span class="math-tex">\(g+c+h-f\)</span> equals</p>
<p style="display:inline">8</p>
<p style="display:inline">29</p>
<p style="display:inline">14</p>
<p style="display:inline">6</p>
Step-by-Step Solution
Key Concept: The locus of points equidistant from two intersecting lines is the pair of angle bisectors, found by equating their perpendicular distance formulas and representing them as a combined second-degree equation.
<p>Locus of point <span class="math-tex">\(P(x, y)\)</span> whose distance from<br />
<span class="math-tex">\(x+2 y+7=0 \& 2 x-y+8=0\)</span> are equal is<br />
<span class="math-tex">\(\Rightarrow \frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}\)</span><br />
Combined equation of lines<br />
<span class="math-tex">\((x-3 y+1)(3 x+y+15)=0\)</span><br />
<span class="math-tex">\(\Rightarrow 3 x^{2}-3 y^{2}-8 x y+18 x-44 y+15=0\)</span><br />
<span class="math-tex">\(\Rightarrow x^{2}-y^{2}-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0\)</span><br />
<span class="math-tex">\(\Rightarrow x^{2}-y^{2}+2 h x y+2 g x 2+2 f y+c=0\)</span><br />
<span class="math-tex">\(h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5\)</span><br />
<span class="math-tex">\(\Rightarrow {g}+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}\)</span><br />
<span class="math-tex">\(=8+6=14\)</span></p>
Correct Answer: C