Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
Cumulative Standards Review

Exercise 28 Page 414

Practice makes perfect
a We are given four points: A(1,1), B(1,5), C(5,5) and D(7,1). These are the vertices of a quadrilateral ABCD. Furthermore, we are told that a trapezoid is a four-sided figure with exactly one pair of parallel sides. We need to decide if ABCD is a trapezoid or not. A graph of the points is shown below.

As we can see, side BC is parallel to side AD. Besides those lines, there is no other pair of parallel sides, which fulfills the given definition of a trapezoid. Therefore, the quadrilateral is indeed a trapezoid.

b We are given four points: A(1,1), B(1,5), C(5,5) and D(7,1). These are the vertices of quadrilateral ABCD. Furthermore, we are ask where should we move point B so that the quadrilateral obtained is a parallelogram — a figure where both pairs of opposite sides are parallel. A graph of the original points is shown below.
As we can see, side BC is already parallel to side AD. We only need to make sure that sides AB and CD are parallel as well. Recall that two lines are parallel when they have the same slope. We can use the Slope Formula. m = y_2-y_1/x_2-x_1Here (x_1,y_1) and (x_2,y_2) are two known points. If we use it with the points C(5,5) and D(7,1) we will know the slope of the line segment CD. Then we will be able to relocate B so line segment AB has the same slope. Let's calculate the slope of CD.
m = y_2-y_1/x_2-x_1
m = 1- 5/7- 5
m = - 4/2
m = - 2
Now we just need to relocate point B to some new coordinates (x,y), such that the slope for AB is - 2 as well. Notice however, that we should not change the current B y-coordinate, or else BC will not be parallel to AD anymore. We can use the Slope Formula with the new B(x,5) point and the known point A(1,1).
m = y_2-y_1/x_2-x_1
m = 1- 5/1- x
m = - 4/1-x
Now let's set that equal to - 2. This way, we can solve for the new B x-coordinate so that AB and CD are parallel.
- 4/1-x = - 2
- 4/1-x(1-x) = - 2 (1-x)
- 4 = - 2 (1-x)
-4 = -2 + 2x
-2 = 2x
2x = - 2
x = -1
We need to relocate point B(1,5) to B(-1,5). Let's do so to verify that we get a new pair of parallel, opposite sides.

We confirm that with the replacement of B(1,5) ⇒ B(-1,5), the line segments AB and CD become parallel while we keep BC and AD parallel as well.