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weqwewe [10]
3 years ago
15

What are the new vertices of parallelogram DEFG if the parallelogram is reflected across the vertical line x = -2, also graphed?

Mathematics
1 answer:
Leni [432]3 years ago
3 0

Answer:

B

Step-by-step explanation:

Since x = - 2 is the line of reflection then the coordinates of the image will be the same distance in the x- direction from x = 2 but on the opposite side.

The y- coordinates of the points remain unchanged.

D is 3 units left of x = - 2 thus D' is 3 units to the right of x = - 2

E is 1 unit left of x = - 2 thus E' is 1 unit to the right of x = - 2

F is 1 unit left of x = - 2 thus F' is 1 unit to the right of x = - 2

G is 3 units left of x = - 2 thus G' is 3 units to the right of x = - 1

Thus

D(- 5, 5 ) → D'(1, 5 )

E(- 3, 3 ) → E'(- 1, 3 )

F(- 3, - 4 ) → F'(- 1, - 4 )

G(- 5, - 2 ) → G'(1, - 2 )

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scoray [572]

Answer18:

The quadrilateral ABCD is not a parallelogram

Answer19:

The quadrilateral ABCD is a parallelogram

Step-by-step explanation:

For question 18:

Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)

The slope of a line is given m=\frac{Y2-Y1}{X2-X1}

Now,

The slope of a line AB:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{6-(-1)}{(-4)-(-4)}

m=\frac{7}{0}

The slope is 90 degree

The slope of a line BC:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{6-6}{(-4)-(-1)}

m=\frac{0}{(-3)}

The slope is zero degree

The slope of a line CD:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{(-4)-6}{2-2}

m=\frac{-10}{0}

The slope is 90 degree

The slope of a line DA:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{(-1)-(-4)}{(-4)-(2)}

m=\frac{3}{-6}

m=\frac{-1}{2}

The slope of the only line AB and CD are the same.

Thus, The quadrilateral ABCD is not a parallelogram

For question 19:

Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)

The slope of a line is given m=\frac{Y2-Y1}{X2-X1}

Now,

The slope of a line AB:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{2-3}{3-(-2)}

m=\frac{-1}{5}

The slope of a line BC:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{(-1)-2}{2-3}

m=\frac{-3}{-1}

m=3

The slope of a line CD:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{0-(-1)}{(-3)-2}

m=\frac{-1}{5}

The slope of a line DA:

m=\frac{Y2-Y1}{X2-X1}

m=\frac{3-0}{(-2)-(-3)}

m=3

The slope of the line AB and CD are the same

The slope of the line BC and DA are the same

Thus, The quadrilateral ABCD is a parallelogram

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≠ ⇒ not equal to

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