<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.

# Parallel and Perpendicular Lines in the Coordinate Plane

## Lines with slopes that are equal or opposite reciprocals of each other.

Estimated6 minsto complete
%
Progress
Practice Parallel and Perpendicular Lines in the Coordinate Plane
Progress
Estimated6 minsto complete
%
Slope of Parallel and Perpendicular Lines

Find the equation of the line parallel to y=2x4\begin{align*}y=2x-4\end{align*} that passes through the point (2,3)\begin{align*}(2,-3)\end{align*}. Then, find the equation of the line perpendicular to y=2x4\begin{align*}y=2x-4\end{align*} that passes through the point (2,3)\begin{align*}(2,-3)\end{align*}. How are the two lines that you found related?

#### Guidance

Consider two lines. There are three ways that the two lines can interact:

1. They are parallel and so they never intersect.
2. They are perpendicular and so they intersect at a right angle.
3. They intersect, but they are not perpendicular.

Recall that the slope of a line is a measure of its steepness. For a line written in the form y=mx+b\begin{align*}y=mx+b\end{align*}, “m\begin{align*}m\end{align*}” is the slope. Given two lines, their slopes can help you to determine whether the lines are parallel, perpendicular, or neither.

In the past you learned that two lines are parallel if and only if they have the same slope. In Example A and Guided Practice #1, you will prove this criteria.

In the past you also learned that two lines are perpendicular if and only if they have slopes that are opposite reciprocals. This means that if the slope of one line is m\begin{align*}m\end{align*}, the slope of a line perpendicular to it will be 1m\begin{align*}-\frac{1}{m}\end{align*}. Another way of thinking about this is that the product of the slopes of perpendicular lines will always be -1. (Note that (m)(1m)=mm=1\begin{align*}(m)\big(-\frac{1}{m}\big)=-\frac{m}{m}=-1\end{align*}). In Examples B and C and Guided Practice #2 and #3, you will prove this criteria.

Example A

Consider two lines y=ax+b\begin{align*}y=ax+b \end{align*} and y=ax+c\begin{align*}y=ax+c\end{align*} with bc\begin{align*}b\ne c\end{align*}. Note that these two lines have the same slope, a\begin{align*}a\end{align*} .

1. How do you know that the two lines are distinct (not the same line?)
2. Use algebra to find the point of intersection of the lines. What happens?

Solution:

1. The two lines are distinct because they have different y\begin{align*}y\end{align*}-intercepts. The first line has a y\begin{align*}y\end{align*}-intercept at (0,b)\begin{align*}(0, b)\end{align*}and the second line has a y\begin{align*}y\end{align*}-intercept at (0,c)\begin{align*}(0,c)\end{align*}.
2. You can use substitution to attempt to find the point of intersection.

y=ax+b\begin{align*}y=ax+b\end{align*} and y=ax+c\begin{align*}y=ax+c\end{align*}

Therefore:

ax+bb=ax+c=c

This is a contradiction because it was stated that bc\begin{align*}b \ne c\end{align*}. Therefore, these two lines do not have a point of intersection. This means the lines must be parallel. This proves that if two lines have the same slope, then they are parallel. You will prove the converse of this statement in Guided Practice #1.

Example B

Consider rectangle ABCD\begin{align*}ABCD\end{align*} with BC=m\begin{align*}BC=m\end{align*}EC=1\begin{align*}EC=1\end{align*} and perpendicular lines AE\begin{align*}\overleftrightarrow{AE}\end{align*} and BE\begin{align*}\overleftrightarrow{BE}\end{align*}.

Find the length of AD\begin{align*}AD\end{align*}. Then, show that ADE\begin{align*}\triangle ADE\end{align*} is similar to ECB\begin{align*}\triangle ECB\end{align*}.

Solution: Because it is a rectangle, AD=BC=m\begin{align*}AD=BC=m\end{align*}. The two triangles are similar because they have congruent angles. Let BEC=θ\begin{align*}\angle BEC =\theta\end{align*} and label all angles in the picture in terms of θ\begin{align*}\theta\end{align*}.

You can see that each of the three triangles in the picture have the same angle measures, so they must all be similar. In particular, ADE\begin{align*}\triangle ADE\end{align*} is similar to ECB\begin{align*}\triangle ECB\end{align*}.

Example C

Use the fact that ADE\begin{align*}\triangle ADE\end{align*} is similar to ECB\begin{align*}\triangle ECB\end{align*} to find the length of DE\begin{align*}DE\end{align*}. Then, find the slopes of lines AE\begin{align*}\overleftrightarrow{AE}\end{align*} and BE\begin{align*}\overleftrightarrow{BE}\end{align*} and show that their product is -1.

Solution: Because ADE\begin{align*}\triangle ADE\end{align*} is similar to ECB\begin{align*}\triangle ECB\end{align*}, the following proportion is true:

m1=DEm

Solving this proportion you have that DE=m2.\begin{align*}DE=m^2.\end{align*}

The slopes of the lines can be found using riserun\begin{align*}\frac{rise}{run}\end{align*}. The slope of line AE\begin{align*}\overleftrightarrow{AE}\end{align*} is mm2=1m\begin{align*}-\frac{m}{m^2}=-\frac{1}{m}\end{align*} and the slope of BE\begin{align*}\overleftrightarrow{BE}\end{align*} is m1=m\begin{align*}\frac{m}{1}=m\end{align*}

The product of the slopes is (1m)(m)=mm=1\begin{align*}\Big(-\frac{1}{m}\Big)(m)=-\frac{m}{m}=-1\end{align*}.

This proves that if two lines are perpendicular, then their slopes will be opposite reciprocals (the product of the slopes will be -1). You will prove the converse of this statement in Guided Practice #2 and #3.

Concept Problem Revisited

To find the equation of the line parallel to \begin{align*}y=2x-4\end{align*} that passes through the point \begin{align*}(2, -3)\end{align*}, remember that parallel lines must have equal slopes. This means that the new line must have a slope of \begin{align*}2\end{align*} and pass through the point \begin{align*}(2,-3)\end{align*}. All you need to do is solve for the \begin{align*}y\end{align*}-intercept.

The equation of the line is \begin{align*}y=2x-7\end{align*}.

To find the equation of the line perpendicular to \begin{align*}y=2x-4\end{align*} that passes through the point \begin{align*}(2,-3)\end{align*}, remember that perpendicular lines will have opposite reciprocal slopes. This means that the new line must have a slope of \begin{align*}-\frac{1}{2}\end{align*} and pass through the point \begin{align*}(2,-3)\end{align*}. Again, all you need to do is solve for the \begin{align*}y\end{align*}-intercept.

The equation of the line is \begin{align*}y=-\frac{1}{2}x-4\end{align*}

The two lines that were found \begin{align*}\left( y=2x-7 \right .\end{align*} and \begin{align*}\left . -\frac{1}{2}x-4 \right )\end{align*} are also perpendicular. Note that they have opposite reciprocal slopes.

#### Vocabulary

The slope of a line is a measure of its steepness. One way to find the slope of a line is to calculate \begin{align*}\frac{rise}{run}\end{align*} or \begin{align*}\frac{\triangle y}{\triangle x}\end{align*} for two points on the line.

Two lines are parallel if they never intersect. Parallel lines have slopes that are equal.

Two lines are perpendicular if they intersect at a right angle \begin{align*}(90^\circ)\end{align*}. Perpendicular lines have slopes that are opposite reciprocals. Also, all vertical lines are perpendicular to all horizontal lines.

#### Guided Practice

1. Consider two parallel lines \begin{align*}y=ax+b\end{align*} and \begin{align*}y=cx+d\end{align*} with \begin{align*}b\ne d\end{align*}. Show that \begin{align*}a=c\end{align*}.

2. Consider two lines intersecting at the origin as shown below. Find the lengths of the legs of each triangle. Then, show that \begin{align*}\triangle BCO\end{align*} is similar to \begin{align*}\triangle ODA\end{align*}.

3. Using the picture from #2, find the slopes of lines \begin{align*}\overleftrightarrow{AO}\end{align*} and \begin{align*}\overleftrightarrow{BO}\end{align*} and verify that their product is -1. Then use the fact that \begin{align*}\triangle BCO\end{align*} is similar to \begin{align*}\triangle ODA\end{align*} to show that \begin{align*}\overleftrightarrow{AO}\end{align*} and \begin{align*}\overleftrightarrow{BO}\end{align*} must be perpendicular.

1. Suppose \begin{align*}a\ne c\end{align*}. You can solve a system of equations to find the point of intersection of the two lines.

\begin{align*}y=ax+b\end{align*} and \begin{align*}y=cx+d\end{align*}

Therefore:

If \begin{align*}a\ne c\end{align*}, then this point exists so the lines intersect. This is a contradiction because it was stated that the lines were parallel. Therefore, \begin{align*}a\end{align*} must be equal to \begin{align*}c\end{align*}. This proves that if two lines are parallel then they must have the same slope.

2. The lengths of the legs of the triangles are shown below.

\begin{align*}\triangle BCO\sim \triangle ODA\end{align*} with a ratio of \begin{align*}m:1\end{align*} by \begin{align*}SAS \sim\end{align*}. \begin{align*}\frac{BC}{OD}=\frac{m}{1}\end{align*} and \begin{align*}\frac{CO}{DA}=\frac{1}{\frac{1}{m}}=\frac{m}{1}.\end{align*} Also \begin{align*}\angle C\cong\angle D\end{align*}.

3. The slope of line \begin{align*}\overleftrightarrow{AO}\end{align*} is \begin{align*}-\frac{1}{m}\end{align*} and the slope of \begin{align*}\overleftrightarrow{BO}\end{align*} is \begin{align*}\frac{m}{1}\end{align*}. The product of the slopes is \begin{align*}\Big(-\frac{1}{m}\Big)(m)=-\frac{m}{m}=-1\end{align*}.

Because \begin{align*}\triangle BCO\end{align*} is similar to \begin{align*}\triangle ODA\end{align*}, their corresponding angles must be congruent. This means that:

• \begin{align*}m\angle OCB=m\angle DOA\end{align*}
• \begin{align*}m\angle BOC=m\angle DAO\end{align*}

Also, because they are right triangles:

• \begin{align*}m\angle OCB +m\angle BOC=90^\circ\end{align*}
• \begin{align*}m\angle DOA+m\angle DAO=90^\circ\end{align*}

By substitution, \begin{align*}m\angle DOA+m\angle BOC=90^\circ \end{align*}. Because \begin{align*}\angle DOA, \angle BOC\end{align*} and \begin{align*}\angle AOB\end{align*} form a straight line, the sum of their measures must be \begin{align*}180^\circ \end{align*}. Therefore, \begin{align*}m\angle AOB\end{align*} must be \begin{align*}90^\circ \end{align*}.

Because \begin{align*}m \angle AOB=90^\circ\end{align*}\begin{align*}\overleftrightarrow{AO}\end{align*} and \begin{align*}\overleftrightarrow{BO}\end{align*} must be perpendicular. This proves that if two lines have opposite reciprocal slopes, then they are perpendicular.

#### Practice

1. Describe the three ways that two lines could interact. Draw a picture of each.

2. What does it mean for two lines to be parallel? How are the slopes of parallel lines related?

3. What does it mean for two lines to be perpendicular? How are the slopes of perpendicular lines related?

4. Use algebra to show why the lines \begin{align*}y=3x-4\end{align*} and \begin{align*}y=3x+7\end{align*} (lines with the same slope) must be parallel.

5. Use the method from Example B and Example C to show why the slopes of lines \begin{align*}\overleftrightarrow{FK}\end{align*} and \begin{align*}\overleftrightarrow{KG}\end{align*} must be opposite reciprocals. Assume that \begin{align*}FGHJ\end{align*} is a rectangle.

6. Find the line parallel to \begin{align*}y=3x-5 \end{align*} that passes through \begin{align*}(2,11).\end{align*}

7. Find the line perpendicular to \begin{align*}y=3x-5\end{align*} that passes through \begin{align*}(6,11).\end{align*}

8. Find the line parallel to \begin{align*}3x+4y=7\end{align*} that passes through \begin{align*}(4,2).\end{align*}

9. Find the line perpendicular to \begin{align*}3x+4y=7\end{align*} that passes through \begin{align*}(3,10).\end{align*}

10. Find the line parallel to \begin{align*}y=5\end{align*} that passes through \begin{align*}(2,16).\end{align*}

11. Find the line perpendicular to \begin{align*}y=5\end{align*} that passes through \begin{align*}(2,16)\end{align*}

12. Find the line parallel to \begin{align*}y=-\frac{1}{3}x-4\end{align*} that passes through \begin{align*}(6,8).\end{align*}

13. Find the line perpendicular to \begin{align*}y=-\frac{1}{3}x-4\end{align*} that passes through \begin{align*}(6,8).\end{align*}

14. Line \begin{align*}a\end{align*} passes through the point \begin{align*}(2,4)\end{align*} and \begin{align*}(3,6)\end{align*}. Line \begin{align*}b\end{align*} passes through the points \begin{align*}(6,7)\end{align*} and \begin{align*}(11,17)\end{align*}. Are lines \begin{align*}a\end{align*} and \begin{align*}b\end{align*} parallel, perpendicular, or neither?

15. Line \begin{align*}a\end{align*} passes through the point \begin{align*}(1,-1)\end{align*} and \begin{align*}(6,14)\end{align*}. Line \begin{align*}b\end{align*} passes through the points \begin{align*}(9,3)\end{align*} and \begin{align*}(-6,8)\end{align*}. Are lines \begin{align*}a\end{align*} and \begin{align*}b\end{align*} parallel, perpendicular, or neither?

### Vocabulary Language: English

Parallel

Parallel

Two or more lines are parallel when they lie in the same plane and never intersect. These lines will always have the same slope.
Perpendicular

Perpendicular

Perpendicular lines are lines that intersect at a $90^{\circ}$ angle. The product of the slopes of two perpendicular lines is -1.
Slope

Slope

Slope is a measure of the steepness of a line. A line can have positive, negative, zero (horizontal), or undefined (vertical) slope. The slope of a line can be found by calculating “rise over run” or “the change in the $y$ over the change in the $x$.” The symbol for slope is $m$