Question
Find the point on X-axis which is equidistant from P(2,-5) and Q(-2,9).
✓
Answer
According to the distance formula, the distance 'd' between two points $(a, b)$ and $(c, d)$ is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}$
As the point is on the $x=$ axis it is of the form $(x, 0)$
Distance from point $P=\sqrt{(2-x)^2+(-5-0)^2}=\sqrt{(2-x)^2+25}$
Distance from point $Q =\sqrt{(-2- x )^2+(9-0)^2}=\sqrt{(2+ x )^2+81}$
As the two points are equidistant from ( $x , 0$ )
$\sqrt{(2-x)^2+25}=\sqrt{(2+x)^2+81}$
Squaring both sides
$(2-x)^2+25=(2+x)^2+81$
Expanding and simplifying
$-4 x+25=4 x+81$
$8 x=-56$
$x=-7$
Hence the point is $(-7,0)$
$d=\sqrt[2]{(a-c)^2+(b-d)^2}$
As the point is on the $x=$ axis it is of the form $(x, 0)$
Distance from point $P=\sqrt{(2-x)^2+(-5-0)^2}=\sqrt{(2-x)^2+25}$
Distance from point $Q =\sqrt{(-2- x )^2+(9-0)^2}=\sqrt{(2+ x )^2+81}$
As the two points are equidistant from ( $x , 0$ )
$\sqrt{(2-x)^2+25}=\sqrt{(2+x)^2+81}$
Squaring both sides
$(2-x)^2+25=(2+x)^2+81$
Expanding and simplifying
$-4 x+25=4 x+81$
$8 x=-56$
$x=-7$
Hence the point is $(-7,0)$
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