If the equation of the locus of point equidistant from the points $\left(a_{1}, b_{1}\right)$ and $\

Question

If the equation of the locus of point equidistant from the points $\left(a_{1}, b_{1}ight)$ and $\left(a_{2}, b_{2}ight)$ is $\left(a_{1}-a_{2}ight) x+\left(b_{1}-b_{2}ight) y+c=0$, then $c=$\

Professor PC Thomas Classes · Answer

(d) : Let $\alpha, \beta$ is the point of locus, equidistant from ( $a_{1}, b_{1}$ ) and ( $a_{2}, b_{2}$ ) is given by\ $\left(\alpha-a_{1}ight)^{2}+\left(\beta-b_{1}ight)^{2}=\left(\alpha-a_{2}ight)^{2}+\left(\beta-b_{2}ight)^{2}$\ $\Rightarrow a_{1}^{2}+b_{1}^{2}-2 a_{1} \alpha-2 b_{1} \beta-a_{2}^{2}-b_{2}^{2}$\ $+2 a_{2} \alpha+2 b_{2} \beta=0$\ $\Rightarrow 2\left(a_{2}-a_{1}ight) \alpha+2\left(b_{2}-b_{1}ight) \beta+a_{1}^{2}$\ $+b_{1}{ }^{2}-b_{2}{ }^{2}-a_{2}{ }^{2}=0$\ $\Rightarrow\left(a_{2}-a_{1}ight) x+\left(b_{2}-b_{1}ight) y+\frac{1}{2}\left(a_{1}^{2}+b_{1}^{2}ight.$\ $\left.-a_{2}{ }^{2}-b_{2}{ }^{2}ight)=0$\ $\Rightarrow c=-\frac{1}{2}\left[a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2}ight]$\ $=\frac{1}{2}\left[a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}ight]$