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The diagonals of quadrilateral EFGH intersect at D(3,3). EFGCH has vertices at E(4,7) and F(-3,5). What must be the coordinates of G and H to ensure that EFGH is a parallelogram.

Sagot :

MrRoyal

Answer:

[tex]G = (2,-1)[/tex]

[tex]H = (9,1)[/tex]

Step-by-step explanation:

Given

[tex]E = (4,7)[/tex]

[tex]F=(-3,5)[/tex]

[tex]D =(3,3)[/tex] --- Diagonal

Required

Determine the coordinates of G and H

Since EFGH is a parallelogram, then:

D is the midpoint of EG and EF

For EG, we have:

[tex](3,3) = \frac{1}{2}(4 + x,7+y)[/tex]

Where x, y are the coordinates of G.

Multiply both sides by 2

[tex](6,6) = (4+x,7+y)[/tex]

By comparison:

[tex]4 + x = 6[/tex] ==> [tex]x =2[/tex]

[tex]7 + y= 6[/tex] ==> [tex]y = -1[/tex]

So:

[tex]G = (2,-1)[/tex]

For FH, we have:

[tex](3,3) = \frac{1}{2}(-3 + x,5+y)[/tex]

Where x, y are the coordinates of H.

Multiply both sides by 2

[tex](6,6) = (-3 + x,5+y)[/tex]

By comparison:

[tex]-3+x = 6[/tex] ==> [tex]x = 9[/tex]

[tex]5 + y = 6[/tex] ==> [tex]y =1[/tex]

So:

[tex]H = (9,1)[/tex]

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