Points (4, 5) and (10, 5) are vertices of a rectangle.

The length of the rectangle is twice the width.

The other vertices are located at (4, ___ ) and (10, ___ ). What could these ordered pairs be?

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Answer:

the remaining two vertices must be (4, -7) and (10, -7).

Step-by-step explanation:

The vertices (4, 5) and (10, 5) lie on the same horizontal line, y = 5.  The length of the rectangle is the difference between the two x-coordinates:  10 - 4, or 6.

Let's denote 6 as the width.  Then the length is twice that, or 12.

Then the remaining two vertices must be (4, -7) and (10, -7).

Check:  is the width 6?  Does 10 - 4 equal 6?  Yes

Is the length twice the width, or 12?  Does 5 - (-7) = 12?  YES

The other vertices are located at (4, -7 ) and (10, -7)

Given two points on a rectangle as (4, 5) and (10, 5)

Since both coordinates lies on the same y axis, then the width of the triangle is expressed as;

width = 2(10-4)

width = 12 units

The y coordinate of the other vertices will be;

y = 5 - 12

y = -7

Hence the other vertices are located at (4, -7 ) and (10, -7)

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