Question

The midpoint of the sides of a triangle are (5,1),(3,-5)and(-5,-1).find the coordinates of the vertices of the triangle ​

Answer 1

Answer:

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Answer 2

Answer:

A(-3,5) ,B(13,-3) ,C(-7,-7)

Step by step explaination:

Let the vertices of triangle ABC be A(x1,y1) ,B(x2,y2) ,C(x3,y3) and given mid points of sides AB, BC, CA are (5,1), (3,-5), (-5,-1) respectively.

Therefore,

x1+x2÷2 = 5 =10....(1)

x2+x3÷2= 3 =6......(2)

x3+x1÷2= -5 =-10...(3)

Adding (1),(2)&(3)

2x1 + 2x2 +2x3 = 6

=x1 + x2 + x3 = 3....(4)

(4)-(2) = x1 =3-6= -3

(4)-(3) = x2=3+10= 13

(4)-(1) = x3=3-10= -7

____________________

y1+y2÷2 = 1 = 2....(5)

y2+y3÷2= -5 = -10......(6)

y3+y1÷2= -1 = -2...(7)

Adding (5),(6)&(7)

2y1 + 2y2 +2y3 = -10

=y1 + y2 + y3 = -5....(8)

(8)-(6) = y1 = -5+10 = 5

(8)-(7) = y2= -5+2 = -3

(8)-(5) = y3= -5-2 = -7

___________________

Therefore ,vertices of the triangle are:

A(-3,5) ,B(13,-3) ,C(-7,-7)

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