McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
4. Parallel and Perpendicular Lines
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Exercise 43 Page 244

Practice makes perfect
a We are given the points A(- 3,3), B(3,5), and C(- 4,0) and are asked to plot them on a coordinate plane. To do so, we have to remember that the first coordinate corresponds to the x-axis and the second to the y-axis.
b We are asked to determine the coordinates of a fourth point D that would form a parallelogram. To accomplish this, we will do three things.
  1. Find the line parallel to AB that passes through C.
  2. Find the line parallel to AC that passes through B.
  3. Find the point of intersection of those lines.

Let's get started!

Line Parallel to AB That Passes Through C

To write the equations for our lines, we will use the slope-intercept form. y= mx+ b In this form m is the slope and b the y-intercept. Remember that two parallel lines have the same slope. We will start with the line parallel to AB. Consider the following diagram.

We can see above that the slope of AB is m= 26 or, simplifying the fraction, m= 13. y= 1/3x+ b To find the y-intercept b, we will use the fact that the line passes through the point C(- 4,0).
y=1/3x+b
0=1/3( - 4)+b
Solve for b
0=- 1/3 * 4+b
0=- 4/3+b
4/3=b
b=4/3
Now that we found the slope m= 13 and y-intercept b= 43, we can write the line parallel to AB that passes through C. y= 1/3x+ 4/3 We have the following diagram.

Line Parallel to AC That Passes Through B

Now, let's find the line parallel to AC that passes through B. Consider the diagram below.

We can see above that the slope is m= 31, which is the same as m= 3. y= 3x+ b. Similar to above, to find the y-intercept b, we will use the fact that the line passes through the point B(3,5).
y=3x+b
5=3* 3+b
Solve for b
5=9+b
- 4=b
b=- 4
Now that we found the slope m= 3 and y-intercept b= -4, we can write the line parallel to AC that passes through B. y= 3x+( -4) We can already see the parallelogram in the diagram!

Finding the Point of Intersection

Finally, we write and label D the point of intersection of the blue and the red lines.

Therefore, the fourth point, the one that would make a parallelogram, is D(2,2).

c We are asked what is the minimum number of points that could be moved to make the parallelogram a rectangle. To do so, we must remember that a rectangle has four right angles. Let's try moving only one point, for example B.

Notice that no matter where we move the point B, ∠ C is not a right angle. Similar logic is true for moving any other single point — the opposite angle is not a right angle. It looks like moving only one point is not enough to construct a rectangle. Let's try moving two points, B and C.

As shown in the diagram above, we can make our parallelogram a rectangle by moving 2 points. Since 1 point is not enough, then 2 is the minimum number of points that could be moved to make our parallelogram a rectangle. Note that we moved opposite points. Following a similar process we could move A and D instead.