Question
Find the point on the x-axis which is equidistant from $(2, -5)$ and $(-2, 9).$
✓
Answer
We know that a point on the x-axis is of the form $(x, 0)$. So, let the point $P(x, 0)$ be equidistant from $A(2, –5)$ and $B(–2, 9)$. Then
$PA = PB$
$\Rightarrow PA^2 = PB^{2}$
$\Rightarrow (2 - x)^2 + (-5 - 0)^2= (-2 - x)^2 + (9 - 0)^2$
$\Rightarrow 4 + x^2 - 4x + 25 = 4 + x^2 + 4x + 81$
$\Rightarrow - 4x + 25 = 4x + 81$
$\Rightarrow 8x = -56$
$\Rightarrow \;x = \frac{{ - 56}}{8} = - 7$
Hence, the required point is (-7, 0)
Check:
$PA = \sqrt {{{\{ 2 - ( - 7)\} }^2} + {{( - 5 - 0)}^2}}$
$= \sqrt {81 + 25} = \sqrt {106}$
$PB = \sqrt {{{\{ - 2 - ( - 7)\} }^2} + {{(9 - 0)}^2}}$
$= \sqrt {25 - 81} = \sqrt {106}$
$\because$
$PA = PB$
$\therefore$ Our solution is checked.
$PA = PB$
$\Rightarrow PA^2 = PB^{2}$
$\Rightarrow (2 - x)^2 + (-5 - 0)^2= (-2 - x)^2 + (9 - 0)^2$
$\Rightarrow 4 + x^2 - 4x + 25 = 4 + x^2 + 4x + 81$
$\Rightarrow - 4x + 25 = 4x + 81$
$\Rightarrow 8x = -56$
$\Rightarrow \;x = \frac{{ - 56}}{8} = - 7$
Hence, the required point is (-7, 0)
Check:
$PA = \sqrt {{{\{ 2 - ( - 7)\} }^2} + {{( - 5 - 0)}^2}}$
$= \sqrt {81 + 25} = \sqrt {106}$
$PB = \sqrt {{{\{ - 2 - ( - 7)\} }^2} + {{(9 - 0)}^2}}$
$= \sqrt {25 - 81} = \sqrt {106}$
$\because$
$PA = PB$
$\therefore$ Our solution is checked.
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