Math, asked by raginisoren14, 11 months ago

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

Answers

Answered by Anonymous
11

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

Answered by jitumahi89
24

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)

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