Question

The vertices of triangle ABC are A(-3,2) B(-1,-4) C(5,2).If M and N are the midpoints of AB and AC respectively ,show that 2MN=BC

Answer 1

M is midpoint of AB

M(x,y) = (-3-1/2 , 2-4/2)

=(-2,-1)

N is the midpoint of AC

N(x,y) = (-3+5/2 , 2+2/2)

= (1,2)

Dist.(MN) = √(1-(-2))^2 + (2-(-1))

= √9+9

=3√2 units

dist.(BC) =√5-(-1)^2 + 2-(-4)^2

=√36+36

=6√2

hence 2MN = BC

Answer 2

Shown that 2MN = BC if The vertices of triangle ABC are A(-3,2) B(-1,-4) C(5,2) , M and N are the midpoints of AB and AC  respectively

Given:

The vertices of triangle ABC are A(-3,2) B(-1,-4) C(5,2)

M and N are the midpoints of AB and AC  respectively

To Find:

Show that 2MN = BC

Solution:

Formulas:

Distance between two points (x₁ , y₁) and (x₂, y₂) is given by:

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

Mid point between two points (x₁ , y₁) and (x₂, y₂) is given by:

( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

M is mid point of A(-3,2) and B(-1,-4)

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

M is mid point of A(-3,2) and C(5,2)

Hence N = (-3 +5)/2 , (2 + 2)/2   =  1 , 2

Distance between B(-1,-4) and C(5,2)

BC = $\sqrt{(5-(-1))^2+(2-(-4))^2} = 6\sqrt{2}$

Distance between M(-2,-1) and N(1,2)

MN = $\sqrt{(1-(-2))^2+(2-(-1))^2} = 3\sqrt{2}$

6√2 = 2 * 3√2

BC = 2 MN

2MN = BC

QED

Hence Proved

Shown that 2MN = BC