Question

12. The vertices of 3ABC are A(1, 8), B(-2, 4), C(8, -5). If M and N are the midpoints
of AB and AC respectively, find the slope of MN and hence verify that MN is
parallel to BC.

Answer 1

The vertices of 3ABC are A(1, 8), B(-2, 4), C(8, -5). If M and N are the midpoints
of AB and AC respectively.
Use midpoint section formula ,
I mean, if A(x,y) is the midpoint of two points (x₁,y₁) and (x₂,y₂),
Then, x = (x₁ + x₂)/2 and y = (y₁ + y₂)/2.

M is the midpoint of A(1,8) and B(-2,4) .
so, M = [(1 -2)/2 , (8 + 4)/2 ] = (-0.5 , 6)

N is the midpoint of A(1,8) and C(8, -5).
so, N = [(1 + 8)/2, (8 -5)/2] = (4.5 , 1.5)

Now, slope of MN = (1.5 - 6)/(4.5 + 0.5) = -4.5/5 = -9/10

Now, we have to find slope of BC to check it is parallel to MN or not.
slope of BC = (-5 - 4)/(8 + 2) = -9/10

Here you can see that slope of MN = slope of BC.
So, MN is parallel to BC.

Answer 2

Solution :

Given A( 1,8) , B( -2,4) ,and

C(8,-5) are three vertices of

∆ABC. M , N are midpoints of

AB and AC respectively .

i ) M is the midpoint of

A( 1,8) = ( x1 , y1 ) ,

B( -2 , 4 ) = ( x2 , y2 )

Coordinates of M

= [ ( x1 + x2 )/2 , ( y1 + y2 )/2 ]

= [ ( 1-2)/2 , ( 8+4 )/2 ]

= ( -1/2 , 6 )

Similarly ,

ii ) coordinates of N = (9/2 , 3/2)

iii ) Slope of a line segment

B( -2 , 4 ) = ( x1 , y1 )

C( 8 , -5) =( x2 , y2 )

m1= ( y2 - y1 )/( x2 - x1 )

m1 = ( -5 - 4 )/( 8 + 2 )

m1 = -9/10 ---( 1 )

iv ) Slope of a line segment MN

m2 = ( 3/2 - 6 )/( 9/2 + 1/2 )

= ( 3 - 12 )/( 9 +1 )

= ( -9 )/10

m2 = -9/10-----( 2 )

From ( 1 ) and ( 2 ) ,

m1 = m2

Therefore ,

MN // BC

••••