If the equation of the locus of a point equidistant from the points (a_1,b_1) and (a_2,b_2…

Question

If the equation of the locus of a point equidistant from the points $(a_1,b_1)$ and $(a_2,b_2)$ is $(a_1-a_2)x+(b_1-b_2)y+c=0$, then the value of $c$ is:

Accepted Answer

Since, the points $\left(a_1, b_1\right)$ and $\left(a_2, b_2\right)$ satisfy the equation. So, that

$$
\begin{aligned}
& a_1\left(a_1-a_2\right)+b_1\left(b_1-b_2\right)+c=0 ~~........(1) \\
& \text { and } a_2\left(a_1-a_2\right)+b_2\left(b_1-b_2\right)+c=0 ~~.........(2) \\
& \text { On adding Eqs. (i) and (ii), we get } \\
& \left(a_1+a_2\right)\left(a_1-a_2\right)+\left(b_1+b_2\right)\left(b_1-b_2\right)+2 c=0 \\
& \Rightarrow  2 c=-\left(a_1^2-a_2^2+b_1^2-b_2^2\right) \\
& \Rightarrow  c=\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right)
\end{aligned}
$$

Solutions · 1