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  • If the equation of the locus of a point equidistant from the points <img alt="\mathrm{\left(a_{1}, b_{1}\right)\: and \: \left(a_{2}, b_{2}\right)\: is \left(a_{1}-a_{2}\right) x+\left(b_{1}-b_{2}\

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

Option: 1

\mathrm{a_{1}^{2}-a_{2}^{2}+b_{1}^{2}-b_{2}^{2}}


Option: 2

\mathrm{\sqrt{a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2}}}


Option: 3

\mathrm{\frac{1}{2}\left(a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2}\right)}


Option: 4

\mathrm{\frac{1}{2}\left(a_{2}^{2}-a_{1}^{2}+b_{2}^{2}-b_{1}^{2}\right)}


Answers (1)

best_answer

Let \mathrm{\left ( h,k \right )} be the point on the locus. Then by the given conditions

\mathrm{\left(h-a_{1}\right)^{2}+\left(k-b_{1}\right)^{2}=\left(h-a_{2}\right)^{2}+\left(k-b_{2}\right)^{2}}

\mathrm{\Rightarrow \quad 2 h\left(a_{12}-a_{2}\right)+2 k\left(b_{1}-b_{2}\right)+a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}=0}
\mathrm{\Rightarrow \quad h\left(a_{1}-a_{2}\right)+k\left(b_{1}+b_{2}\right)+\frac{1}{2}\left(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}\right)=0\quad \cdots(i)}

Also, since \mathrm{(h, k)} lies on the given locus, therefore

\mathrm{\left(a_{1}-a_{2}\right) h+\left(b_{1}-b_{2}\right) k+c=0\quad \cdots(ii)}

Comparing (i) and (ii), we get
\mathrm{c=\frac{1}{2}\left(a_{2}^{2}+b_{2}^{2}--a_{1}^{2}-b_{1}^{2}\right)}
 

Posted by

Pankaj

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