Question

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

• A. $\frac{1}{2}\left(a^{2}_{2}+b^{2}_{2}-a^{2}_{1}-b^{2}_{1}\right)$
• B. $a^2_1+a^2_2-b^2_1-b^2_2$
• C. $\frac{1}{2}\left(a^{2}_{1}+a^{2}_{2}-b^{2}_{1}-b^{2}_{2}\right)$
• D. $\sqrt{a^{2}_{1}+b^{2}_{2}-a^{2}_{2}-b^{2}_{2}}$

Answer 1

Answer: (A)

Let $p (x, y)$
$(x − a_1)^2 + (y − b_1)^2 = (x − a_2)^2 + (y − b_2)^2$
$\left(a_{1}-a_{2}\right)x+\left(b_{1}-b_{2}\right)y+\frac{1}{2}\left(b^{2}_{2}-b^{2}_{1}+a^{2}_{2}-a^{2}_{1}\right)=0.$
Hence, (A) is the correct answer.