Question

find the point on x axis which are equidistant from points p (-3,4)&q(1,-4)​

Answer 1

$\large\underline{\sf{Solution-}}$

Let the point on x - axis be (x, 0) which is equidistant from the points A (- 3, 4) and B (1, - 4).

We know,

Distance between two points A and B is given by

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

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

Now,

According to statement,

$\rm :\longmapsto\:PA = PB$

On squaring both sides, we get

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

$\rm :\longmapsto\: {(x + 3)}^{2} + {(0- 4)}^{2} = {(x - 1)}^{2} + {(0 + 4)}^{2}$

$\rm :\longmapsto\: {x}^{2} + 9 + 6x + 16 = {x}^{2} + 1 - 2x + 16$

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

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

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

$\bf\implies \:x = - 1$

Hence,

$\bf:\longmapsto\:The \: point \: on \: x - axis \: be \: P \: (-1,0)$

Additional Information :-

Section Formula

Section Formula is used to find the co ordinates of the point(Q) Which divides the line segment joining the points (B) and (C) internally in the ratio m : n

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

Midpoint Formula :-

Midpoint Formula is used to find the midpoint of line segment joining two points

${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{x_2 + x_1}{2},  \dfrac{y_2 + y_1}{2}\Bigg) \quad}}}}$