Question

Find a point on the y-axis equidistant from (-5, 2) and (9,-2).​

Answer 1

$\bold \red \star \: \large \underline{Solution:-}$

°•° Let A and B is two points which equidistant from the point P respectively.

$\dot\: \: A( - 5 ,2) \\ \\ </p><p></p><p> \dot\: \: P(0,y) \\ \\ </p><p></p><p> \dot\: \: B(9, - 2)$

$or, \: AP=BP.$

$\rule{200}3$

Let's apply section formula,

$\sqrt{{(x_2 - x_1) }^{2} +{( y_2 - y_1)}^{2} } = \sqrt{{(x_2 - x_1) }^{2} +{( y_2 - y_1)}^{2} }. \\ \\ \sqrt{{(0 + 5) }^{2} +{( y - 2)}^{2} } = \sqrt{{(9 - 0) }^{2} +{( 2 + y)}^{2} }.$

$Square \: Both \: sides.$

$25 + \cancel{ {y}^{2}} + \cancel 4 - 4y = 81 + \cancel 4 + \cancel{ {y}^{2}} + 4y. \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 8y = 56. \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = - \frac{ \cancel{56}}{ \cancel8} . \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = 7.$

$\rule{200}3$

We know that distance never be negative.

therefore,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: |y = 7.|$