Question

If the equation of the locus of 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 $c=$?

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

Answer 1

Answer: (D)

$\frac{1}{2}(a{{}_2}^2+b{{}_2}^2-a{{}_1}^2-b{{}_1}^2)$

The locus of the point equidistant from two given points is the perpendicular bisector of the segment joining the two segments.
The equation of the perpendicular bisector is given as -
$(a_1-a_2)x+(b_1-b_2)y+c=0$
The midpoint of the segment joining the two points $(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2})$ lies on this line. On substituting the coordinates of the midpoint in the equation of the line we get,
$(a_1-a_2)(a_1+a_2)+(b_1-b_2)(b_1+b_2)=-2c$
On simplifying this expression, we get option D as the answer.