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# Find a point which is equidistant from the points $ \mathrm{A}(-5,4) $ and $ \mathrm{B}(-1,6) $ ? How many such points are there?

Given points are $A (-5, 4)$ and $B (-1, 6)$.

To do:

We have to find a point that is equidistant from the point $A (-5, 4)$ and $B (-1, 6)$ and the number of such points.

Solution:

Let $P(x, y)$ be the point which is equidistant from the points $A (-5, 4)$ and $B (-1, 6)$.

This implies,

$PA = PB$

Squaring on both sides, we get,

\( \Rightarrow(\mathrm{PA})^{2}=(\mathrm{PB})^{2} \)

\( \Rightarrow(-5-x)^{2}+(4-y)^{2}=(-1-x)^{2}+(6-y)^{2} \)

\( \Rightarrow 25+x^{2}+10 x+16+y^{2}-8 y=1+x^{2}+2 x+ 36+y^{2}-12 y \)

\( \Rightarrow 25+10 x+16-8 y=1+2 x+36-12 y \)

\( \Rightarrow 8 x+4 y+41-37=0 \)

\( \Rightarrow 8x+4y+4=0 \)

\( \Rightarrow 2x+y+1=0 \)....(i)

Mid-point of \( \mathrm{AB}=\left(\frac{-5-1}{2}, \frac{4+6}{2}\right) \)

\( =(-3,5) \)

From (i),

At point $(-3,5)$,

$2(-3)+5+1=0$

The mid-point of $AB$ satisfies (i).

Hence, infinite number of points, in fact all points which are solutions of the equation $2x + y + 1 = 0$ are equidistant from the points $A$ and $B$.

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