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*}. Then, 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*}. How are the two lines that you found related?

### Slope of Parallel and Perpendicular Lines

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

- They are parallel and so they never intersect.
- They are perpendicular and so they intersect at a right angle.
- 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 \begin{align*}y=mx+b\end{align*}, “\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 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 \begin{align*}m\end{align*}, the slope of a line perpendicular to it will be \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 \begin{align*}(m)\big(-\frac{1}{m}\big)=-\frac{m}{m}=-1\end{align*}).

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

How do you know that the two lines are distinct (not the same line?)

The two lines are distinct because they have different \begin{align*}y\end{align*}-intercepts. The first line has a \begin{align*}y\end{align*}-intercept at \begin{align*}(0, b)\end{align*}and the second line has a \begin{align*}y\end{align*}-intercept at \begin{align*}(0,c)\end{align*}.

Use algebra to find the point of intersection of the lines. What happens?

You can use substitution to attempt to find the point of intersection.

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

Therefore:

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

This is a contradiction because it was stated that \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.*

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

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

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

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

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

\begin{align*}\frac{m}{1}=\frac{DE}{m}\end{align*}

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

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

The product of the slopes is \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).*

### Examples

#### Example 1

Earlier, you were asked how were the two lines that you found related.

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.

\begin{align*}-3 &=2(2)+b\\ -3 &=4+b\\ b &=-7\\\end{align*}

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.

\begin{align*}-3 &=-\frac{1}{2}(2)+b\\ -3 &=-1+b\\ -4 &=b\end{align*}

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.

#### Example 2

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*}.

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:

\begin{align*}ax+b &=cx+d\\ x(a-c) &=d-b\\ x &=\frac{d-b}{a-c}\end{align*}

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. *

#### Example 3

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*}.

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*}.

#### Example 4

Using the picture from #3, 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.

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.*

### Review

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?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 10.4.