Question

if the point (m,n) is equidistant from (2,3) And (6-1), prove that m-n=3​

Answer 1

Basic Concept Used :-

Distance Formula :-

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

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

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

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

Let's solve the problem now!!

Let us suppose that coordinates are represented as

• P(m, n)

• A(2, 3)

• B(6, - 1)

According to statement,

• Point P is equidistant from A and B,

$\bf\implies \:PA = PB$

• On squaring both sides, we get

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

$\rm :\longmapsto\: {(m - 2)}^{2} + {(n - 3)}^{2} = {(m - 6)}^{2} + {(n + 1)}^{2}$

$\rm :\longmapsto\: {(m - 2)}^{2} - {(m - 6)}^{2} = {(n + 1)}^{2} - {(n - 3)}^{2}$

$\rm :\longmapsto\:(m - 2 + m - 6)(m - 2 - m + 6) = (n + 1 - n + 3)(n + 1 + n - 3)$

$\rm :\longmapsto\:(2m - 8)(4) = (2n - 2)(4)$

$\rm :\implies\:2m - 8 = 2n - 2$

$\rm :\implies\:2m - 2n = 6$

$\bf\implies \:m - n = 3$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

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

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

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

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

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

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

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

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