lf 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 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_{1}^{2} - a_{2}^{2} - b_{2}^{2}}

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Solution

The correct option is B $\frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2})$
Let $P (h, k)$ be a point which is equidistant from $A (a_{1}, b_{1})$ and $B (a_{2}, b_{2})$
i.e. $P A = P B$
$\Rightarrow (h - a_{1})^{2} + (k - b_{1})^{2} = (h - a_{2})^{2} + (k - b_{2})^{2}$
$\Rightarrow 2 (a_{1} - a_{2}) h + 2 (b_{1} - b_{2}) k + (a_{2}^{2} - a_{1}^{2}) + (b_{2}^{2} - b_{1}^{2}) = 0$
$\Rightarrow (a_{1} - a_{2}) x + (b_{1} - b_{2}) y + \frac{1}{2} (a_{2}^{2} - a_{1}^{2}) + \frac{1}{2} (b_{2}^{2} - b_{1}^{2}) = 0$
Comparing with the given equation of locus
$(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$
$\Rightarrow c = \frac{1}{2} (a_{2}^{2} - a_{1}^{2} + b_{2}^{2} - b_{1}^{2}) = 0$

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Similar questions

(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

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

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

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

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