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 what is the value of c?
(a). ${a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2$
(b). $\sqrt {{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}$
(c). $\dfrac{1}{2}({a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2)$
(d). $\dfrac{1}{2}({a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2)$

Answer 1

Hint: The distance between two points is given by the formula $\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$. Use this to equate the locus of the points equidistant from the given points and then simplify the expression to find the value of c.

Complete step-by-step answer:
It is given that the equation of the locus of a point equidistant from the points $({a_1},{b_1})$ and $({a_2},{b_2})$ is $({a_1} - {a_2})x + ({b_1} - {b_2})y + c = 0$. We need to find the value of c.
We know that the distance between two points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by the formula as follows:
$D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$
Let the point that is equidistant from the two points $({a_1},{b_1})$ and $({a_2},{b_2})$ be (x, y).
The distance between the point (x, y) and the point $({a_1},{b_1})$ is given as follows:
$D = \sqrt {{{({a_1} - x)}^2} + {{({b_1} - y)}^2}} .............(1)$
The distance between the point (x, y) and the point $({a_2},{b_2})$ is given as follows:
$D = \sqrt {{{({a_2} - x)}^2} + {{({b_2} - y)}^2}} .............(2)$
It is given that the distances are the same, we equate equation (1) and (2) as follows:
$\sqrt {{{({a_1} - x)}^2} + {{({b_1} - y)}^2}} = \sqrt {{{({a_2} - x)}^2} + {{({b_2} - y)}^2}}$
Taking square on both sides, we have:
${({a_1} - x)^2} + {({b_1} - y)^2} = {({a_2} - x)^2} + {({b_2} - y)^2}$
We now, expand the squares to get as follows:
${a_1}^2 - 2{a_1}x + {x^2} + {b_1}^2 - 2{b_1}y + {y^2} = {a_2}^2 - 2{a_2}x + {x^2} + {b_2}^2 - 2{b_2}y + {y^2}$
Canceling the common terms, we have:
${a_1}^2 - 2{a_1}x + {b_1}^2 - 2{b_1}y = {a_2}^2 - 2{a_2}x + {b_2}^2 - 2{b_2}y$
Taking all the terms to the right-hand side of the equation, we have:
$0 = - {a_1}^2 + 2{a_1}x - {b_1}^2 + 2{b_1}y + {a_2}^2 - 2{a_2}x + {b_2}^2 - 2{b_2}y$
Grouping the common terms, we have:
$2({a_1} - {a_2})x + 2({b_1} - {b_2})y + - {a_1}^2 - {b_1}^2 + {a_2}^2 + {b_2}^2 = 0$
Dividing the entire equation by 2, we get:
$({a_1} - {a_2})x + ({b_1} - {b_2})y + \left( {\dfrac{{ - {a_1}^2 - {b_1}^2 + {a_2}^2 + {b_2}^2}}{2}} \right) = 0$
Comparing this equation with the given equation $({a_1} - {a_2})x + ({b_1} - {b_2})y + c = 0$, we obtain the value of c as follows:
$c = \dfrac{{{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2}}{2}$
Hence, the correct answer is option (d).

Note: There is a high possibility of making a mistake when shifting the equations between the left-hand and the right-hand side of the equation which may also result in option (c), which is a wrong answer.