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the coordinates for a rhombus are given as (2a,0),(0,2b),(-2a,0) and (0,-2b) write a plan to prove that the midpoints of the sides of a rhombus determine a rectangle using coordinate geometry be sure to include the formulas

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boffeemadrid

Answer:

Step-by-step explanation:

Let the points of rhombus be A(2a,0), B(0,2b), C(-2a,0) and D(0,-2b).

Let midpoints of  AB be P, then

[tex]P=(\frac{0+2a}{2},\frac{0+2b}{2})=(a,b)[/tex]

Let the mid point of BC be Q, then

[tex]Q=(\frac{-2a+0}{2},\frac{0+2b}{2})=(-a,b)[/tex]

Let the mid point of CD be R, then

[tex]R=(\frac{0-2a}{2},\frac{-2b+0}{2})=(-a,-b)[/tex]

and Let the mid point of DA be S, then

[tex]S=(\frac{2a+0}{2},\frac{0-2b}{2})=(a,-b)[/tex]

It can be seen that P lies in quadrant I, Q in Quadrant II, R in III and S in IV, Further P and Q are the reflections of each other in y-axis, Q and R are the reflections of each other in x-axis,R  and  S  are reflection of each other in  y -axis and  S  and  P are reflection of each other in  x -axis.

Hence, the mid points of the rhombus  forms the rectangle.

Ver imagen boffeemadrid
MrRoyal

The coordinates of the rectangle are (a,b), (-a,b) (-a,-b) and (a,-b)

The coordinate of the rhombus are given as: (2a,0), (0,2b), (-2a,0) and (0,-2b)

We start by calculating the midpoint of consecutive points using the following midpoint formula.

[tex]\mathbf{(x,y) = \frac{1}{2}(x_1 + x_2, y_1 + y_2)}[/tex]

For (2a,0) and (0,2b), we have:

[tex]\mathbf{(x,y) = \frac{1}{2}(2a + 0,0+2b)}[/tex]

[tex]\mathbf{(x,y) = \frac{1}{2}(2a,2b)}[/tex]

Divide

[tex]\mathbf{(x,y) = (a,b)}[/tex]

For (0,2b) and (-2a,0), we have:

[tex]\mathbf{(x,y) = \frac{1}{2}(0 - 2a,2b + 0)}[/tex]

[tex]\mathbf{(x,y) = \frac{1}{2}(-2a,2b)}[/tex]

Divide

[tex]\mathbf{(x,y) =(-a,b)}[/tex]

For (-2a,0) and (0,-2b), we have:

[tex]\mathbf{(x,y) = \frac{1}{2}(-2a + 0,0-2b)}[/tex]

[tex]\mathbf{(x,y) = \frac{1}{2}(-2a ,-2b)}[/tex]

Divide

[tex]\mathbf{(x,y) = (-a ,-b)}[/tex]

For (0,-2b) and (2a,0), we have:

[tex]\mathbf{(x,y) = \frac{1}{2}(0 + 2a,-2b+0)}[/tex]

[tex]\mathbf{(x,y) = \frac{1}{2}(2a,-2b)}[/tex]

Divide

[tex]\mathbf{(x,y) = (a,-b)}[/tex]

Hence, the coordinates of the rectangle are (a,b), (-a,b) (-a,-b) and (a,-b)

Read more about rhombus and midpoints at:

https://brainly.com/question/12796015

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