Question

if the point p(x,y) is equidistant from the two points A(-3,2) and B(4,-5). prove that y = x - 2

Answer 1

I hope it helps u......

Answer 2

Answer:

Step-by-step explanation:

Given : Point p(x,y) is equidistant from the two points A(-3,2) and B(4,-5).

To Find : Prove that y = x - 2

Solution :

Calculate the distance between Point p and Point A using distance formula .

Distance formula: $d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^2}$

Point p : $(x_1,y_1) = (x,y)$

Point A : $(x_2,y_2) = (-3,2)$

So, $d=\sqrt{(-3-x)^{2}+(2-y)^2}$  --1

Calculate the distance between Point p and Point B using distance formula .

Distance formula: $d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^2}$

Point p : $(x_1,y_1) = (x,y)$

Point B : $(x_2,y_2) = (4,-5)$

So, $d=\sqrt{(4-x)^{2}+(-5-y)^2}$   --2

Since we are given that Point p(x,y) is equidistant from the two points A(-3,2) and B(4,-5).

So, equate 1 and 2

$\sqrt{(-3-x)^{2}+(2-y)^2}=\sqrt{(4-x)^{2}+(-5-y)^2}$

$\sqrt{9+x^2+6x+4+y^2-4y}=\sqrt{16+x^2-8x+25+y^2+10y}$

$9+x^2+6x+4+y^2-4y=16+x^2-8x+25+y^2+10y$

$6x-4y+13=-8x+10y+41$

$6x+8x=10y+4y+41-13$

$14x=14y+28$

$x=y+2$

$y=x-2$

Hence proved