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.