Question

find the point on the x-axis which is equidistant from point A ( -3,4) and B ( 1,4)​

Answer 1

Basic Concept Used :-

Distance Formula :-

This Formula is used to find the distance between two given points.

Let us assume a line segment joining the points A and B, then distance between A and B is given by

$\sf \: \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)$

Let's solve the problem now!!

• Let suppose that the point on x - axis be P (x, 0)

Given coordinates are

• A (- 3, 4)

and

• B (1, 4)

According to statement,

• Point P is equidistant from A and B,

$\bf :\implies\:PA = PB$

On squaring both sides, we get

$\rm :\longmapsto\: {PA}^{2} = {PB}^{2}$

$\rm :\longmapsto\: {(x + 3)}^{2} + \cancel{{(y - 4)}^{2}} = {(x - 1)}^{2} + \cancel{(y - 4)}^{2}$

$\rm :\longmapsto\: \cancel{x}^{2} + 9 - 6x = \cancel{x}^{2} + 1 + 2x$

$\rm :\longmapsto\: - 6x - 2x = 1 - 9$

$\rm :\longmapsto\: - 8x = - 8$

$\bf\implies \:x = 1$

$\bf:\implies\:\: Point \: on \: x - axis \: be \: P \: (1, \: 0)$

Additional Information :-

$\underline{\bigstar\:\textsf{Section Formula\; :}}$

Section Formula is used to find the co ordinates of the point C (x, y) which divides the line segment joining the points (A) and (B) internally in the ratio m : n, then

${\underline{\boxed{\sf{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n}, \: \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)$

$\underline{\bigstar\:\textsf{Mid Point Formula\; :}}$

Mid Point formula is used to find the Mid points on any line.

Let us assume a line segment joining the points A and B and let C be the midpoint of AB, then coordinates of C is

${\underline{\boxed{\sf{\quad \bigg(\dfrac{x_1 + x_2}{2} \; ,\; \dfrac{y_1 + y_2}{2} \bigg) \quad}}}}$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)$