Question

Find the midpoints of the sides of a triangle whose vertices are A(1,-1),B(4,-1),C(4,3)

Answer 1

Step-by-step explanation:

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

Mid pt of AB = 5/2 and -1

mid pt of BC = 4 and 1

mid pt of AC = 5/2 and 1

hope it helps you buddy

Answer 2

Answer:

Mid point of AB is $(\frac{5}{2},-1)$

Mid point of BC is $(4,1)$

Mid point of CA is $(\frac{5}{2} ,1)$

Step-by-step explanation:

since we know that mid point is given as $(\frac{x_{1}+x_{2} }{2},\frac{y_{1}+y_{2}}{2})$.

for side AB we have

$x_{1}=1,x_{2}=4, y_{1} =-1 , and,y_{2}=-1$

apply formula we get,

$(\frac{1+4}{2},\frac{-1+-1}{2})$

$(\frac{5}{2},-1)$

So mid point of AB is $(\frac{5}{2},-1)$

for side BC we have

$x_{1}=4,x_{2}=4, y_{1} =-1 , and,y_{2}=3$

apply formula we get,

$)(\frac{4+4}{2},\frac{-1+3}{2}$)

$(4,1)$

So mid point of BC is $(4,1)$

for side CA we have

$x_{1}=4,x_{2}=1, y_{1} =3 , and,y_{2}=-1$

apply formula we get,

($\frac{4+1}{2},\frac{3+-1}{2}$)

$(\frac{5}{2} ,1)$

So mid point of CA is $(\frac{5}{2} ,1)$