Question

Find the point on x-axis which is equidistant from the points (2,-2) and (-4,2)

Answer 1

Let the given points be

• A (2, -2)
• B (-4, 2)

$\:$

The point which is equidistant from points A and B is on x axis. So let the required point be P (x, 0)

$\:$

Now, We know that

Point P is equidistant from points A and B

$\:$

So,

AP = PB

Applying distance formula,

$AP = \sqrt{(x - 2) {}^{2} + (0 - ( - 2)) {}^{2} }$

$\leadsto \: \sf{AP = \sqrt{x {}^{2} + 4 - 4x + 4} }$

$\to \: \sf{AP = \sqrt{ {x}^{2} - 4x + 8} }$

For finding distance BP:

$\sf{BP = \sqrt{(x - ( - 4) {}^{2} + (0 - 2) ^{2} } }$

$\dashrightarrow \: \sf{BP = \sqrt{(x + 4) ^{2} + {( - 2)}^{2} } }$

$\dashrightarrow \: \sf{BP = \sqrt{ {x }^{2} + 16 + 8x + 4 } }$

$\implies \sf{BP = \sqrt{ {x}^{2} + 8x + 20 } }$

$\tt{Since \: AP = BP}$

$\sf{\sqrt{ {x}^{2} - 4x + 8 } = \sqrt{ {x}^{2} + 8x + 20 } }$

$\tt{Squaring \: on \: both \: sides,}$

$\sf{{x}^{2} - 4x + 8 = {x}^{2} + 8x + 20}$

$\dashrightarrow \: \sf{ 8 - 4x = 8x + 20}$

$\dashrightarrow \: \sf{ - 8x - 4x = 20 - 8}$

$\dashrightarrow \: \sf{ - 12x = 12}$

$\implies \sf{x = \dfrac{ - 12}{12} }$

$\therefore \boxed{ \bf{x = - 1}}$

So, the required point on x axis is

P (-1, 0)