Consider the line $L : a x + b y + c = 0$ and the points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ such that $(a x_{1} + b y_{1} + c) (a x_{2} + b y_{2} + c) < 0,$ then the possible number of points on the line L which are equidistant from the points A and B is :

0

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1

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infinitely many

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nothing can be said

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Solution

The correct options are
A $1$
B infinitely many
D $0$
Given line $L : a x + b y + c = 0$
Also, $(a x_{1} + b y_{1} + c) (a x_{2} + b y_{2} + c) < 0$
$\Rightarrow (x_{1}, y_{1}), (x_{2}, y_{2})$ lies on the opposite side of $L .$
Also, $P A = P B$
${P A}^{2} = {P B}^{2}$
Clearly, $P$ has to lie on the perpendicular bisector of $A B .$ (Say $L$ )
Hence, there are three cases possible
$L$ coincides with given line => infinitely many points
$L$ is parallel to given line => no such points
$L$ intersects the given line at one point
Hence, A,B,C are correct options.

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