Algebraic Proofs - Connecting Algebra and Geometry - High School Geometry Unlocked (2016)

High School Geometry Unlocked (2016)

Chapter 6. Connecting Algebra and Geometry

Lesson 6.4. Algebraic Proofs

In this lesson, you’ll learn more about proofs and how to use algebra to complete proofs. When you are instructed to complete an algebraic proof, it means that you are expected to rely almost entirely on algebraic expressions and equations, and not a specific example with numerical values. The idea is usually that you’re constructing a proof that works for any scenario, not just with a certain set of numbers. As a reminder, there are usually several different possible ways that you can set up a proof. There is no “right” or “wrong” approach, as long as you follow the instructions set by the task, and show all of your steps from beginning to end.

Your friend, a sculpture artist, is making a sketch of her next project which will use mirrors and lasers. She draws two parallel lines, which represent two large mirrors. She’s certain that these lines are parallel, because she measured a constant distance between them. However, she’s not sure if that means that the lines will have the same slope. She needs to be certain of this, in order to achieve the desired effect with the lasers. She asks you to explain it to her algebraically, so that she knows her sketch will work.

Can you prove algebraically that two parallel lines will always have the same slope?

One way to complete this proof is to use similar triangles. This works because we’ll be able to use the legs of the triangles to represent the slope. Construct both a horizontal and vertical transversal intersecting the lines, as shown:

The transversals form triangles with our parallel lines, and we can prove that these triangles have corresponding, congruent angles. Using the horizontal transversal, we know that cdoabo. And from the vertical transversal, we know that ∠dco ≈ ∠bao. Both of those pairs are alternate interior angles. Additionally, ∠cod ≈ ∠aob, because these represent the same angle shared by both triangles.

Since the two triangles have three pairs of corresponding, congruent angles, we know that the triangles are similar.

Now, consider what we know about the slopes of these lines. The two triangles are right triangles, since the horizontal and vertical transversals necessarily intersect each other at a right angle. Therefore, we can use the legs of the right triangles to represent the rise and run of the lines.

And using the definition of similar triangles, we know that . Thus, it is proved that the two slopes are equal.

Given:

l1l2

m1 is the slope of l1

m2 is the slope of l2

Prove:

m1 × m2 = −1

To review rotations,
see Chapter 1
of this book.

One way to complete this proof is to use the rules for rotations of coordinates. This works because any point on l1 would be a 90° rotation of a corresponding point on l2, and vice versa.

First, we’ll say that we’ve translated l1 and l2 so that their intersection is at the origin. We know that any time we translate a line, the slope will stay the same. Draw a figure with two perpendicular lines which intersect at the origin:

Choose a point on l2 and call it (a, b). If we rotate this point 90° about the origin, the rotated coordinate will be (−b, a).

This rotated coordinate is sure to lie on line l2, because we know that l1 and l2 are perpendicular, and thus they meet at right angles.

Now, since each of these lines has expressions for two coordinates (including the origin), we can calculate their slopes.

Slope of l1:

Slope of l2:

Now, find that m1 × m2 = −1 :

Simplify.

To set up an algebraic proof for a geometric figure, ask yourself what you know about the figure in terms of rules and equations. For example, what is true about the equations of parallel lines? Perpendicular lines? What features do certain shapes have? You should be able to find a rule or fact that is directly related to the fact that you are trying to prove.

In the figure above, lines m, n, p, and q have the equations (y = 3x + 4), (3y = −x + 21), ( = x + 4), and (y = −x + 1), respectively.

Is the shaded quadrilateral, which is bound by m, n, p, and q, a rectangle?

First, consider the definition of a rectangle—it is a figure with four sides that meet at right angles. Therefore, we can prove that the figure is a rectangle by proving that the adjacent sides meet at right angles.

One way to prove that the adjacent sides are perpendicular to each other is to identify their slopes. We have the equations for these lines, so that makes comparing the slopes easy. First, make sure that each of the equations is in the form of y = mx + b.

The equation for line m is already in standard form:

y = 3x + 4

The equation for line n should be put into standard form:

3y = −x + 21

Divide both sides of the equation by 3.

Simplify.

The equation for line p should be put into standard form:

= x + 4

y = 3(x + 4)

Multiply both sides of the equation by 3.

y = 3x + 12

Simplify.

The equation for line q is already in standard form:

y = −x + 1

Now that the equations are in standard form, we can compare their slopes. Lines m and p both have a slope of 3, while lines n and q both have a slope of −. Perpendicular lines have opposite reciprocal slopes, which is what we see here.

Line m is perpendicular to lines n and p, line n is perpendicular to m and q, and so on. Therefore, we can be sure that the figure is a rectangle, because its vertices are 90° angles.

In the coordinate plane, a quadrilateral is bound by the lines m (3y = 6x + 12), n (y = x + 9), p ( = x), and q (y − 2 = 5x).

Is the quadrilateral a parallelogram?

We know that a parallelogram is a quadrilateral in which opposite sides are parallel and congruent. We can try to prove that opposite sides are parallel, or that they are congruent—either would be sufficient to prove that the quadrilateral is a parallelogram. Note that this works for parallelograms since one rule can’t be true without the other; when working with other shapes, however, you may have to prove multiple facts together.

Since no coordinates were given, it will be easier to prove whether opposite sides are parallel. Put each equation into the form of y = mx + b; then you can compare the slopes.

Line m:

3y = 6x + 12

y =

Divide both sides of the equation by 3.

y = 2x + 4

Simplify.

Line n:

y = x + 9

Already in standard form.

Line p:

= x

y − 9 = 2x

Multiply both sides of the equation by 2.

y = 2x + 9

Add 9 to both sides of the equation.

Line q:

y − 2 = 5x

y = 5x + 2

Add 2 to both sides of the equation.

Compare the slopes of the lines. Lines m and p must be parallel, since they both have a slope of 2. However, lines n and q have different slopes ( and 5), so they are not parallel. The figure is not a parallelogram, but rather a trapezoid.

How would you go
about proving whether
the opposite sides
are congruent?

A triangle is bound by the lines (y = x + 2), (y = x + 2), and (y = x − 3). Find the perimeter of the triangle.

To find the perimeter of the triangle, we will need to find the lengths of the sides. Since we only have equations given, a good first step would be to find the intersection points of the lines. These intersection points will be the vertices of our triangle.

Find the intersection of y = x + 2 and y = x + 2. Set the expressions equal to each other:

x + 2 = x + 2

x = x

Subtract 2 from both sides of the equation.

xx = 0

Subtract x from both sides of the equation.

x = 0

Simplify.

x = 0

Divide both sides of the equation by .

The x-coordinate of the intersection is 0. Plug x = 0 into one of the two equations to solve for y:

y = x + 2

y = 0 + 2

y = 2

Therefore, the intersection of y = x + 2 and y = x + 2 is (0, 2).

Can you find the remaining two intersections? Try to calculate them on your own. Then read on for the coordinates.

The intersection of (y = x + 2) and (y = x − 3) is (10, 12).

The intersection of (y = x + 2) and (y = x − 3) is (6, 6).

Next, we’ll need to calculate the distance between each pair of points. Use the distance formula.

Find the distance from (0, 2) to (10, 12).

The distance from (0, 2) to (10, 12) is , or ≈ 14.14.

Find the distance from (0, 2) to (6, 6).

The distance from (0, 2) to (6, 6) is , or ≈ 7.21.

Find the distance from (6, 6) to (10, 12).

The distance from (6, 6) to (10, 12) is , or ≈ 7.21.

Finally, add the three lengths to find the perimeter:

+ +

≈ 14.14 + 7.21 + 7.21

≈ 28.56

DRILL

CHAPTER 6 PRACTICE QUESTIONS

Directions: Complete the following problems as specified by each question. For extra practice after answering each question, try using an alternative method to solve the problem or check your work.

Click here to download a PDF of Chapter 6 Practice Questions.

1.Line p is represented by the equation 3y + 2x = 9. At what point will line p intersect the line that contains points (0, 2) and (5, 8)?

2.What is the area of a triangle with vertices (2, 1), (−5, 1), and (−5, −9)?

3.On line segment JKL, JK and KL have a ratio of 1:3. If point J is at (0, 3) and point K is at (3, 7), what is the total distance of JKL?

4.Write the equation of a parabola with focus (−3, 9) and directrix x = 0.

5.A parabola has an equation of y = (x + 3)2 + 2. What is the equation of the horizontal line that passes through the parabola’s vertex?

6.A triangle is bound by the lines 3y + 4x = −11, 3y − 4x = 29, and x = −2. Is this a right triangle? Prove why or why not.

7.Is the triangle described in question 6 an isosceles triangle? Prove why or why not.

8.A bakery opened and quickly grew in popularity. Later, however, the popularity reached a peak and ultimately died down to the point that the bakery had to close. The bakery’s success can be represented with the equation y = −(x)2 + 5 with x representing the length of time in years since the bakery opened and y representing the number of customers per month, in thousands, where x and y are both positive values. After how many years did the bakery’s popularity peak and what was the maximum number of customers?

9.Aiden lives 2 miles east and 3 miles north of a park. Mia lives 1 mile east and 4 miles south of the same park. Aiden and Mia want to meet exactly in the middle of their two homes. How many miles will each person have to travel, and what is the relationship between that point and the park?

SOLUTIONS TO CHAPTER 6 PRACTICE QUESTIONS

1.(15/28, 37/14)

First, find the equation of the second line. The first step is to calculate the slope using the formula . In this case, the formula is . So far, the equation is y = (6/5)x + b. Now, calculate b by substituting the x- and y-values from one of the points: 8 = 6/5(5) + b, then 8 = 6 + b, so b = 2. The full equation is y = 6/5x + 2.

Next, rewrite the other equation in the same form. First, 3y = −2x + 9. Then, divide by 3 to get y = (−2/3)x + 3.

Now, set the two equations equal to each other: (6/5)x + 2 = −2/3x + 3. To make the math easier, multiply the whole equation by a common denominator, which is 15. The result is 18x + 30 = −10x + 45. Now, combine like terms to get 28x = 15, so x = 15/28. Next, plug that into one of the equations to find the y-value: y = 6/5(15/28) + 2, then y = 90/140 + 2 = 9/14 + 2 = 37/14.

The coordinate is (15/28, 37/14).

2.35

First, draw the triangle to make the problem easier. You can tell that this is a right triangle, because the legs are parallel to the axes (one leg lies on the line y = 1, and the other leg lies on the line x = −5). This makes the math easier—you could use the distance formula, but in this case it’s easiest to just count. The base goes from −5 to 2 on the x-axis, which is a total distance of 7. The height goes from 1 to −9 on the y-axis, for a total distance of 10. Plug those numbers into the area formula A = (1/2)bh to get A = 1/2(7)(10) = 35.

3.20

First, find the distance between (0, 3) and (3, 7). Using a right triangle, the horizontal distance is 3 and the vertical distance is 4, so the hypotenuse is 5 (3:4:5 triangle). You can also use the distance formula to get the same result. This distance is 5, and that is the first part of the 1:3 ratio, which means the other part, KL, measures 3 times as much, or 15. The total distance is 5 + 15 = 20.

4.x = (−1/6)(y − 9)2 − 3/2

Use the formula for a horizontal parabola: x = a(yk)2 + h. The vertex must be halfway between the directrix and the focus. Drawing the figure can help to visualize this. The middle of 0 and −3 is −1.5, so the vertex is at (−1.5, 9). The vertex is (h, k), so now the equation is x = a(y − 9)2 − 1.5. Since 1/(4a) is the distance between the vertex and the focus (−1.5), set 1/(4a) = −1.5, then 6a = −1 so a = −1/6. Keep in mind it must be negative for the parabola to face left, which it does based on the directrix and focus. The final equation is x = (−1/6)(y − 9)2 − 3/2.

5.y = 2

First find the vertex. The vertex is (h, k), so based on the equation provided it is (−3, 2). The line must pass through that point. Since it is a horizontal line, it will be in the y = format. The y-value of the vertex is 2, so the line is y = 2.

6.No.

In order for the triangle to be a right triangle, two sides must be perpendicular, in which case the lines’ slopes are the negative reciprocals of each other. For x = −2, a vertical line, a perpendicular line would have to be a horizontal line, which neither of the other two are. So just check the other two to compare their slopes. First, put them into slope-intercept form. The first one is 3y + 4x = −11, so then 3y = −4x − 11, and y = −4/3x − 11/3. If the other line is perpendicular, it must have a slope of 3/4. The other one is 3y − 4x = 29, so then 3y = 4x + 29 and y = 4/3x + 29/3. This line is not perpendicular to the other because its slope is 4/3, not 3/4. Another way of thinking about it is the products of the slopes must equal −1 and in this case the product is −16/9, not −1. Thus, this is not a right triangle.

An alternative method would be to find the vertices (the intersection points of the lines), calculate the distances between them, and then see if the distances satisfy the Pythagorean theorem. But that would be more work for this problem.

7.Yes.

An isosceles triangle has two equal sides. Find the distances between each pair of vertices to determine whether two of the distances are the same. First, find the vertices, which are the points at which the lines intersect. Start with y = −4/3x − 11/3 and y = 4/3x+ 29/3 (as we found in question 6) and set them equal to each other: − 4/3x − 11/3 = 4/3x + 29/3. Multiply both sides by 3: −4x − 11 = 4x + 29. Combine like terms: −40 = 8x, so x = −5. Plug that into one of the equations: y = − 4/3(−5) − 11/3, then y = 20/3 − 11/3 = 9/3 = 3. One point is (−5, 3).

Now, combine y = −4/3x − 11/3 and x = −2. The x-value is −2, so plug that in to find y: y = − 4/3(−2) − 11/3 = 8/3 − 11/3 = − 3/3 = −1. A second vertex is (−2, −1). Next, do the same to find the last vertex with y = 4/3x + 29/3 and x = −2. y = 4/3(−2) + 29/3 = − 8/3 + 29/3 = 21/3 = 7. The last vertex is (−2, 7).

Next, use the distance formula or right triangles to find the distances between the points (−5, 3), (−2, −1), and (−2, 7). For (−5, 3) and (−2, −1), the difference between the x-values is 3 and between the y-values is 4, so this is a 3:4:5 triangle. Their distance is 5. For (−2, −1) and (−2, 7), the distance is just the difference between the y-values, which is 8. For (−2, 7) and (−5, 3), the difference between the x-values is 3, between the y-values is 4, so again this is a 3:4:5 and their distance is 5. Thus, this is an isosceles triangle because two sides have an equal distance of 5.

8.About 2.2 years and 5,000 customers.

Any time you’re asked to find the maximum or minimum value for a parabola, you’re being asked to find the vertex. The equation is in vertex form, so the vertex is (h, k), or (, 5) (approximately (2.2, 5)). The x represents the amount of time in years, so that is 2.2 years, and the y represents the number of customers in thousands, so 5,000 customers.

9.About 3.5 miles per person; 3/2 mile east and 1/2 mile south of the park.

Think of the park like the origin on a coordinate grid. Aiden’s point is (2, 3) and Mia’s point is (1, −4). The question is asking for the midpoint, which is found by taking the average of the x-values and theaverage of the y-values. Here it is: , , or (3/2, −1/2). This means that the meeting spot is 3/2 mile east of the park and 1/2 mile south of the park. Now determine how far each person has to travel. This involves the distance between the two points, which can be found using the distance formula or a right triangle. The difference between the x-values is 1 and the difference between the y-values is 7, so the formula is 12 + 72 = c2 or 50 = c2, so c = , which is about 7.1. They will each travel half that distance, so about 3.5 miles per person.

REFLECT

Congratulations on completing Chapter 6!

Here’s what we just covered.

Rate your confidence in your ability to:


•Understand the standard form for the equation of a line

•Find an equation from a line, and plot a line from its equation

•Find the intersection of two lines

•Calculate the distance or midpoint between two coordinates

•Divide a segment to a specified ratio (e.g., 2:3 or 1:5)

•Find the vertex, roots, and axis of symmetry of a parabola

•Recognize the standard form and vertex form for the equation of a parabola

•Find the focus and directrix of a parabola from its equation

•Find the equation of a parabola given its zeroes, or its focus and directrix

•Prove geometric facts and theorems using algebra

If you rated any of these topics lower than you’d like, consider reviewing the corresponding lesson before moving on, especially if you found yourself unable to correctly answer one of the related end-of-chapter questions.

Access your online student tools for a handy, printable list of Key Points for this chapter. These can be helpful for retaining what you’ve learned as you continue to explore these topics.