Question

A (-3,2) B (3,2) C (-3,-2) are the vertices of the right triangle , right angled at A . Show that the midpoint of the hypotenuse is equidistant form the vertices.

Answer 1

Answer:

Step-by-step explanation:

Formula used:

Midpoint formula:

The midpoint of the line joining$(x_1,y_1) and (x_2,y_2) is\\\\(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Distance formula:

The distance between two points$(x_1,y_1) and (x_2,y_2) is\\\\d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Let S be the midpoint of hypotenuse BC.

Then, S is

$(\frac{3-3}{2},\frac{2-2}{2})$

(0,0)

Now,

$SA=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\SA=\sqrt{(0+3)^2+(0-2)^2}\\SA=\sqrt{9+4}\\SA=\sqrt{13}\\\\SB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\SB=\sqrt{(0-3)^2+(0-2)^2}\\SB=\sqrt{9+4}\\SB=\sqrt{13}\\\\SC=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\SC=\sqrt{(0+3)^2+(0+2)^2}\\SC=\sqrt{9+4}\\SC=\sqrt{13}$

This implies

SA=SB=SC

Hence S is equidistant from A, B and C

Answer 2

Answer:

mid point of BC=(x1+x2/2,y1+y2/2)

BC=(3-3/2,2-2/2)

BC=(0/2,0/2)

BC=(0,0).

distance formula apply in following

dis OA=√(0-(-3))²+(0-2)²

OA=√(3)²+(-29²)

OA=√9+4

OA=√13.

dis OB=√(0-3))²+(0-2)²

OB=√(-3)²+(-2)²

OB=√9+4

OB=√13.

dis OC=√(0-(-3))²+(0-(-2))²

OC=√(3)²+(2)²

OC=√9+4

OC=√13.

Thus,OA=OB=OC.

hence O is equidistant from A,B,C.

Hope it's help.

plz thanks for me.