Question

The vertices of an triangle abc area A(-3,2) B(-1,-4)and C(-5,2).if M and N are the midpoint of AB and AC respectively,show that 2MN is equal to BC​

Answer 1

Answer:

BC = 2 MN

Step-by-step explanation:

Formula used:

The midpoint of the line joining $(x_1,y_1)\:and\:(x_2,y_2)$ is

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

The distance between two points $(x_1,y_1)\:and\:(x_2,y_2)$ is

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

Given points are

A(-3,2) B(-1, -4)and C(-5,2)

Midpoint of AB is M

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\\\M=(\frac{-3-1}{2},\frac{2-4}{2})\\\\M=(\frac{-4}{2},\frac{-2}{2})\\\\M=(-2,-1)$

Midpoint of AC is N

$N=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\\\N=(\frac{-3-5}{2},\frac{2+2}{2})\\\\N=(\frac{-8}{2},\frac{4}{2})\\\\N=(-4,2)$

$Now\\\\BC=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\BC=\sqrt{(-1+5)^2+(-4-2)^2}\\\\BC=\sqrt{(4)^2+(-6)^2}\\\\BC=\sqrt{16+36}\\\\BC=\sqrt{52}\\\\BC=\sqrt{4*13}\\\\BC=2\sqrt{13}.........(1)$

$MN=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\MN=\sqrt{(-2+4)^2+(-1-2)^2}\\\\MN=\sqrt{(2)^2+(-3)^2}\\\\MN=\sqrt{4+9}\\\\MN=\sqrt{13}............(2)$

From (1) and (2)

BC= 2 MN

Answer 2

Answer:

Proved BC = 2MN

Step-by-step explanation:

The vertices of an triangle abc area A(-3,2) B(-1,-4)and C(-5,2).if M and N are the midpoint of AB and AC respectively,show that 2MN is equal to BC​

A(-3,2) B(-1,-4)C(-5,2)

M mid point of AB

M = ( -3 + (-1))/2  , (2 + (-4))/2 = ( -2 , -1)

N mid point of AC

N = ( -3 + (-5))/2  , (2 + 2)/2 = ( -4 , 2)

MN = √( (-2 -(-4))² + (-1 -2)²) = √ (4 + 9) =√13

BC= √( (-5 -(-1))² + (2 -(-4))²) = √(16 + 36) = √52 = 2√13

BC = 2MN

QED