# Lesson 14

Coordinate Proof

• Let’s use coordinates to prove theorems and to compute perimeter and area.

### 14.1: Which One Doesn’t Belong: Coordinate Quadrilaterals

Which one doesn’t belong?

A quadrilateral has vertices $$(0,0), (4,3), (13,\text-9),$$ and $$(9,\text-12)$$.

1. What type of quadrilateral is it? Explain or show your reasoning.
2. Find the perimeter of this quadrilateral.
3. Find the area of this quadrilateral.

1. A parallelogram has vertices $$(0,0), (5,1), (2,3)$$, and $$(7,4)$$. Find the area of this parallelogram.
2. Consider a general parallelogram with vertices $$(0,0),(a,b),(c,d),$$ and $$(a+c,b+d),$$ where $$a,b,c,$$ and $$d$$ are positive. Write an expression for its area in terms of $$a,b,c,$$ and $$d$$.

### 14.3: Circular Logic

Use the applet to answer the questions.

1. First, observe and describe the image:
1. What kind of line is $$BC$$ in reference to the circle?
2. What does the measurement of angle $$BDC$$ appear to be?
2. Move point $$D$$ around. How does the measurement of angle $$BDC$$ appear to change? Use a corner of the index card to compare the size of the angle as it moves.
3. Move point $$D$$ to a location with integer coordinates. Each student in the group should choose a different spot for $$D$$. Calculate the slopes of segments $$BC$$ and $$BD$$. What do your results tell you?
4. Now select the button labeled “slopes.” This will show the slopes of $$BD$$ and $$BC$$ and their product. Move point $$D$$ around. What do you notice about the product of the slopes?
5. Based on the results, write a conjecture that captures what you are seeing.

### Summary

What kind of shape is quadrilateral $$ABCD$$? It looks like it might be a rhombus. To check, we can calculate the length of each side. Using the Pythagorean Theorem, we find that the lengths of segments $$AB$$ and $$CD$$ are $$\sqrt{45}$$ units, and the lengths of segments $$BC$$ and $$DA$$ are $$\sqrt{37}$$ units. All side lengths are between 6 and 7 units long, but they are not exactly the same. So our calculations show that $$ABCD$$ is not really a rhombus, even though at first glance we might think it is.

We did just show that two pairs of opposite sides of $$ABCD$$ are congruent. This means that $$ABCD$$ must be a parallelogram. Checking slopes confirms this. Sides $$AB$$ and $$CD$$ each have slope $$\frac12$$. Sides $$BC$$ and $$DA$$ each have slope 6.

Can we find the area of triangle $$EFG$$? That seems tricky, because we don’t know the height of the triangle using $$EG$$ as the base. However, angle $$EFG$$ seems like it could be a right angle. In that case, we could use sides $$EF$$ and $$FG$$ as the base and height.

To see if $$EFG$$ is a right angle, we can calculate slopes. The slope of $$EF$$ is $$\frac86$$ or $$\frac43$$, and the slope of $$FG$$ is $$\text-\frac{3}{4}$$. Since the slopes are opposite reciprocals, the segments are perpendicular and angle $$EFG$$ is indeed a right angle. This means that we can think of $$EF$$ as the base and $$FG$$ as the height. The length of $$EF$$ is 10 units and the length of $$FG$$ is 5 units. So the area of triangle $$EFG$$ is 25 square units because $$\frac12 \boldcdot 10 \boldcdot 5 = 25$$.

### Glossary Entries

• opposite

Two numbers are opposites of each other if they are the same distance from 0 on the number line, but on opposite sides.

The opposite of 3 is -3 and the opposite of -5 is 5.

• point-slope form

The form of an equation for a line with slope $$m$$ through the point $$(h,k)$$. Point-slope form is usually written as $$y-k = m(x-h)$$. It can also be written as $$y = k + m(x-h)$$.

If $$p$$ is a rational number that is not zero, then the reciprocal of $$p$$ is the number $$\frac{1}{p}$$.