If the equation of the locus of a point equidistant from the points $(a_{1}, b_{1}) a n d (a_{2}, b_{2}) i s (a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$
then the value of c is

\frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2})

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a_{1}^{2} + a_{2}^{2} + b_{1}^{2} - b_{2}^{2}

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\frac{1}{2} (a_{1}^{2} + a_{2}^{2} - b_{1}^{2} - b_{2}^{2})

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\sqrt{a_{1}^{2} + b_{2}^{2} - a_{2}^{2} - b_{2}^{2}}

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Solution

The correct option is A $\frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2})$
Assume the general point as P(h,k), then the locus is
$(h - a_{1})^{2} + (k - b_{1})^{2} = (h - a_{2})^{2} + (k - b_{2})^{2} \Rightarrow (a_{1} - a_{2}) h + (b_{1} - b_{2}) k + \frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2}) = 0 Comparing it with given locus, \Rightarrow c = \frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2})$

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Similar questions

a_{1} x + b_{1} y + c_{1} z + d_{1} = 0

a_{2} x + b_{2} y + c_{2} z + d_{2} = 0

{[{(a_{1} {+ a}_{2} + . . . . a_{n})}^{2} {(b_{1} + b_{2} + . . . . + b_{n})}^{2}]}^{1 / 2} < \sqrt{{a_{1}}^{2} + {b_{1}}^{2} +} \sqrt{{a_{2}}^{2} + {b_{2}}^{2}} + . . . . + \sqrt{{a_{n}}^{2} + {b_{n}}^{2}}

a_{r}, b_{r} (r = 1, 2, ., n)

\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}}

θ

a_{2} x + b_{2} y + c_{2} z = d

\frac{x^{2}}{a_{1}^{2}} - \frac{y^{2}}{b_{1}^{2}} = 1

\frac{x^{2}}{a_{2}^{2}} - \frac{y^{2}}{b_{2}^{2}} = 1

(a_{1}^{2} + a_{2}^{2} + \dots \dots a_{n}^{2}) (b_{1}^{2} + b_{2}^{2} + \dots \dots + b_{n}^{2}) \geq (a_{1} b_{1} + a_{2} b_{2} + \dots \dots + a_{n} b_{n})^{2}

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