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Question 4
May use your own graph paper for this question. Show all algebraic work for full By graphing, find the coordinates of the vertices of the original triangle H L K, give midsegment triangle MNP are M(-1,3), N(4,0), P(-2,-2).

Question 4 May Use Your Own Graph Paper For This Question Show All Algebraic Work For Full By Graphing Find The Coordinates Of The Vertices Of The Original Tria class=

Sagot :

With the mid segment of the triangle MNP as M (-1, 3), N (4, 0), P (-2, -2) the coordinates of the triangle are

  • (-7, 1), (5, 5), and (3, -5)

How to find the coordinates of the triangle when the mid segment is known

The coordinates is calculated from the midsegment using the formula

X= (x₂ + x₁)/2

Y = (y₂ + y₁)/2

Vertex 1 and 2: point  M  (-1, 3),  

X

-1 = (x₂ + x₁)/2

-2 = x₂ + x₁

Y

3 = (y₂ + y₁)/2

6 = y₂ + y₁

Vertex 2 and 3:  point N (4, 0)

X

4 = (x₂ + x₃)/2

8 = x₂ + x₃

Y

0 = (y₂ + y₃)/2

0 = y₂ + y₃

Vertex 1 and 3:  point P(-2, -2)

X

-2 = (x₃ + x₁)/2

-4 = x₃ + x₁

Y

-2 = (y₃ + y₁)/2

-4 = y₃ + y₁

The equations

X

-2 = x₂ + x₁

8 = x₂ + x₃

-4 = x₃ + x₁

Y

6 = y₂ + y₁

0 = y₂ + y₃

-4 = y₃ + y₁

from -2 = x₂ + x₁, we find x₁ = -2 - x₂

plugging in x₁ = -2 - x₂ to -4 = x₃ + x₁ gives

-4 = x₃ + -2 - x₂

-2 = x₃ - x₂

then we have

8 = x₂ + x₃

-2 = x₃ - x₂

solving simultaneously

x₃ = 3, x₂ = 5 then x₁ = -2 - x₂ = -7

from 0 = y₂ + y₃, we find that y₂ = -y₃

plugging in y₂ = -y₃ into 6 = y₂ + y₁ gives

6 = -y₃ + y₁

then we have

-4 = y₃ + y₁

6 = -y₃ + y₁

solving simultaneously gives

y₁ = 1, y₃ = -5, then y₂ = -y₃ = -(-4) = 5

The vertex of the triangle are

(-7, 1), (5, 5), and (3, -5)

Learn more about midpoint segment here:

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