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5.) The diagonals of a quadrilateral intersect at (-1,4). One of the sides of the quadrilateral is bounded by (2,7) and (-3,5).

Part A: Determine the coordinates of the other side in order for the
quadrilateral to be a parallelogram.

Part B: Determine the coordinates of the other side in order for the
quadrilateral to be a square.

5 The Diagonals Of A Quadrilateral Intersect At 14 One Of The Sides Of The Quadrilateral Is Bounded By 27 And 35 Part A Determine The Coordinates Of The Other S class=

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xero099

a) The points of the other side of the parallelogram are [tex](1, 3)[/tex] and [tex](-4,1)[/tex].

b) The points of the other side of the square are [tex](-1,0)[/tex] and [tex](4,2)[/tex].

How to find missing points of quadrilaterals

In this question we shall use vector operations and Pythagorean theorem to determine the location of the missing two points of each quadrilateral.

a) Let be [tex]A(x,y) = (-3, 5)[/tex] and [tex]B(x,y) = (2,7)[/tex] the endpoints of the line segment and [tex]C(x,y) = (-1, 4)[/tex] the point of intersection of the diagonals. The  remaining points of the parallelogram can be found by the following two formulas:

[tex]D(x,y) = A(x,y) + 2\cdot [C(x,y)-A(x,y)][/tex] (1)

[tex]E(x,y) = B(x,y) + 2\cdot [C(x,y)-B(x,y)][/tex] (2)

([tex]A(x,y) = (-3, 5)[/tex], [tex]B(x,y) = (2,7)[/tex], [tex]C(x,y) = (-1, 4)[/tex])

[tex]D(x,y) = (-3,5)+2\cdot (2,-1)[/tex]

[tex]D(x,y) = (1,3)[/tex]

[tex]E(x,y) = (2, 7) + 2\cdot (-3,-3)[/tex]

[tex]E(x,y) = (-4,1)[/tex]

The points of the other side of the parallelogram are [tex](1, 3)[/tex] and [tex](-4,1)[/tex]. [tex]\blacksquare[/tex]

b) Let be [tex]A(x,y) = (-3, 5)[/tex] and [tex]B(x,y) = (2,7)[/tex] the endpoints of the line segment, the coordinates of the missing points are found by the following two formulas:

[tex]D(x,y) = A(x,y) + (AB_{x}\cdot \cos \theta - AB_{y}\cdot \sin \theta, AB_{x}\cdot \sin \theta + AB_{y}\cdot \cos \theta)[/tex] (3)

[tex]E(x,y) = B(x,y) + \overrightarrow{AD}[/tex] (4)

Where [tex]\theta[/tex] is the rotation angle, in degrees.

([tex]A(x,y) = (-3, 5)[/tex], [tex]B(x,y) = (2,7)[/tex], [tex]\theta = -90^{\circ}[/tex])

[tex]\overrightarrow{AB} = (2+3, 7-5)[/tex]

[tex]\overrightarrow{AB} = (5,2)[/tex]

[tex]D(x,y) = (-3, 5) + (5\cdot \cos (-90^{\circ})-2\cdot \sin (-90^{\circ}), 5\cdot \sin (-90^{\circ})+2\cdot \cos (-90^{\circ}))[/tex]

[tex]D(x,y) = (-1, 0)[/tex]

[tex]\overrightarrow{AD} = (-1 + 3, 0-5)[/tex]

[tex]\overrightarrow{AD} = (2, -5)[/tex]

[tex]E(x,y) = (2,7) +(2, -5)[/tex]

[tex]E(x,y) = (4, 2)[/tex]

The points of the other side of the square are [tex](-1,0)[/tex] and [tex](4,2)[/tex]. [tex]\blacksquare[/tex]

To learn more on quadrilaterals, we kindly invite to check this verified question: https://brainly.com/question/25240753

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