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Lesson Explainer: Equations of Parallel and Perpendicular Lines Mathematics

In this explainer, we will learn how to write the equation of a line parallel or perpendicular to another line.

Parallel lines are straight lines that never intersect. To understand the link between parallel lines and their slopes, let us consider two lines of equations 𝑦=π‘šπ‘₯+π‘οŠ§οŠ§ and 𝑦=π‘šπ‘₯+𝑏.

We can solve for the point of intersection by setting the expressions for 𝑦 to be equal: π‘šπ‘₯+𝑏=π‘šπ‘₯+𝑏.

Now, we rearrange so that the π‘₯-terms are on the same side of the equation: π‘šπ‘₯βˆ’π‘šπ‘₯=π‘βˆ’π‘.

By factoring, we find (π‘šβˆ’π‘š)π‘₯=π‘βˆ’π‘.

Finally, we note that we can isolate the variable π‘₯ by dividing by π‘šβˆ’π‘šοŠ§οŠ¨, and so, only if π‘šβ‰ π‘šοŠ§οŠ¨, this gives π‘₯=π‘βˆ’π‘π‘šβˆ’π‘š.

This shows us that two lines intersect only if their slopes are not equal. Conversely, we can conclude that lines of the same slope do not intersect; a pair of parallel lines will have the same slopes.

We can see this by considering the slopes of a pair of parallel lines.

For every π‘₯ units we go across on either line, we must travel 𝑦 units in the vertical direction on both lines; otherwise, the lines will intersect. It is worth noting that there is one small problem with this reasoning, which is if we have vertical lines. In this case, we cannot talk about the slope, since vertical lines do not have a slope. However, we can note that a pair of distinct vertical lines will be parallel. We have the following result.

Property: Slopes of Parallel Lines

If two nonvertical lines are parallel, then they have the same slopes.

If two distinct lines have the same slopes (π‘š=π‘š) or are both vertical, then they are parallel.

This allows us to check if two lines are parallel. For example, consider the lines 𝑦=βˆ’3π‘₯+2 and 3π‘₯+𝑦=1. We can recall that a line given in the form 𝑦=π‘šπ‘₯+𝑐 has a slope of π‘š and a 𝑦-intercept of 𝑐. The first line is given in this form, so its slope is given by the coefficient of π‘₯, which is βˆ’3. We can subtract 3π‘₯ from both sides of the equation of the second line to get 𝑦=βˆ’3π‘₯+1.

The coefficient of π‘₯ is βˆ’3, so its slope is also βˆ’3. It is important to check that the 𝑦-intercepts of the two lines are different. If they had the same 𝑦-intercept, then the two lines would actually be the same line (called coincident lines).

Since the two lines are distinct and have the same slopes, we can conclude that they are parallel. This property allows us to check if any two lines are parallel.

We can now ask the question of how to check if two lines are perpendicular. We can do this by sketching any nonvertical lines, including a right triangle representing its slope.

A line perpendicular to this line will meet the line at right angles. We can find any line like this by rotating the red line 90∘. If we rotate the line 90∘, we will also rotate the triangle 90∘.

We can then calculate the slope of the perpendicular line: for every 𝑏 units we move across, we move π‘Ž units down. Its slope is βˆ’π‘Žπ‘. Thus, if a line has slope π‘π‘Ž, then the lines perpendicular to this line have slope βˆ’π‘Žπ‘. We can think of this in two ways: either we take the negative of the reciprocal of the slope or the product of the slopes as βˆ’1. The same is true in reverse: if two lines have slopes whose product is βˆ’1, then they are perpendicular. We have the following property.

Property: Slopes of Perpendicular Lines

If two nonvertical lines are perpendicular, then their slopes are the negative of the reciprocal of each other. Alternatively, the product of their slopes gives βˆ’1.

If two lines have slopes π‘šοŠ§ and π‘šοŠ¨ such that π‘š=βˆ’1π‘šοŠ§οŠ¨, then they are perpendicular.

The only caveat to this property is if we have horizontal or vertical lines. In this case, we can note that horizontal and vertical lines are perpendicular to each other even though their slopes do not satisfy this property.

In our first example, we will determine the relationship between a pair of lines from their equations.

Example 1: Identifying Whether Two Lines Are Parallel, Perpendicular, or Otherwise

How would you describe the relation between the lines 𝑦=17π‘₯+9 and βˆ’π‘₯+7𝑦+4=0?

  1. Parallel
  2. Coincident
  3. Perpendicular
  4. Intersecting and not perpendicular

Answer

To determine if the lines are parallel or perpendicular, we first want to find their slopes. We can do this by writing both equations in the form 𝑦=π‘šπ‘¦+𝑐, where π‘š is the slope and 𝑐 is the 𝑦-intercept. The first equation is already in this form, so it has slope 17. We can rearrange the second equation to get βˆ’π‘₯+7𝑦+4=07𝑦=π‘₯βˆ’4𝑦=17π‘₯βˆ’47.

Therefore, the second line also has a slope of 17. Since both lines have the same slopes, they are either parallel or coincident. We can see that the first line has a 𝑦-intercept of 9 but the second line has a 𝑦-intercept of βˆ’47. Since the two lines pass through different 𝑦-intercepts, they cannot be the same line, so they are not coincident.

Hence, since the lines have the same slopes and are not coincident, they must be parallel which is option A.

In our next example, we will find the equation of a line given a point on the line and the equation of a parallel line.

Example 2: Finding the Equation of a Line given a Point on the Line and a Parallel Line

Write, in the form 𝑦=π‘šπ‘₯+𝑐, the equation of the line through (βˆ’1,βˆ’1) that is parallel to the line βˆ’6π‘₯βˆ’π‘¦+4=0.

Answer

We first recall that the equation of a line with a slope of π‘š that passes through (π‘₯,𝑦) can be written as π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯). As we have the coordinates of a point on the line, we now need to find its slope. To do this, we use the fact that the line is parallel to the line βˆ’6π‘₯βˆ’π‘¦+4=0. As parallel lines have the same slopes (unless they are both vertical), we want to find the slope of the line βˆ’6π‘₯βˆ’π‘¦+4=0. We can do this by rewriting its equation in the form 𝑦=π‘šπ‘₯+𝑐.

We add 𝑦 to both sides of the equation to get 𝑦=βˆ’6π‘₯+4.

The coefficient of π‘₯ is βˆ’6, so the slope of this line, and thus of our line, is βˆ’6.

Substituting π‘š=βˆ’6, π‘₯=βˆ’1, and 𝑦=βˆ’1 into the equation π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯) gives us π‘¦βˆ’(βˆ’1)=βˆ’6(π‘₯βˆ’(βˆ’1))𝑦+1=βˆ’6(π‘₯+1).

We can now expand the brackets and rearrange to get 𝑦+1=βˆ’6π‘₯βˆ’6𝑦=βˆ’6π‘₯βˆ’7.

In our next example, we will find the equation of a line given a point on the line and two points on another parallel line.

Example 3: Finding the Equation of a Line given a Point on the Line and Two Points on Another Parallel Line

Find, in slope–intercept form, the equation of the straight line passing through the point (3,1) and parallel to the straight line passing through the two points (1,βˆ’1) and (4,βˆ’3).

Answer

We first recall that the slope–intercept form of a line is the equation 𝑦=π‘šπ‘₯+𝑐, where the line has a slope of π‘š and a 𝑦-intercept of 𝑐. We are not given the slope or 𝑦-intercept of this line.

Instead, we are given a point on the line and two points on a parallel line. We can recall that for the lines to be parallel, they need to have the same slopes. We can determine the slope of a line through (π‘₯,𝑦) and (π‘₯,𝑦) by using the formula π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.

Substituting π‘₯=1, 𝑦=βˆ’1, π‘₯=4, and 𝑦=βˆ’3 into the formula for the slope yields π‘š=(βˆ’1)βˆ’(βˆ’3)(1)βˆ’(4)=βˆ’23.

Therefore, the slope of our line is βˆ’23.

Now, we can write the equation of our line in the form π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯), where π‘š is its slope and (π‘₯,𝑦) are the coordinates of a point it passes through. Substituting π‘š=βˆ’23, π‘₯=3, and 𝑦=1 into this equation gives us π‘¦βˆ’1=βˆ’23(π‘₯βˆ’3).

The last step is to rearrange this equation into the form 𝑦=π‘šπ‘₯+𝑐. We start by expanding the brackets to get π‘¦βˆ’1=βˆ’23π‘₯+ο€Όβˆ’23(βˆ’3)π‘¦βˆ’1=βˆ’23π‘₯+2.

We now add 1 to both sides of the equation to obtain 𝑦=βˆ’23π‘₯+3.

In our next example, we will find the equation of a line given a point on the line and the equation of a perpendicular line.

Example 4: Finding the Equation of a Line given a Point on the Line and a Perpendicular Line

Find, in slope–intercept form, the equation of the line perpendicular to 𝑦=2π‘₯βˆ’4 that passes through the point 𝐴(3,βˆ’3).

Answer

We first recall that the slope–intercept form of a line is the equation 𝑦=π‘šπ‘₯+𝑐, where the line has a slope of π‘š and a 𝑦-intercept of 𝑐. We are not given the slope or 𝑦-intercept of this line.

Instead, we are given a point on the line and the equation of a perpendicular line. We can determine the slope of the line by noting that it is perpendicular to the line 𝑦=2π‘₯βˆ’4; perpendicular lines have slopes that multiply to give βˆ’1 (unless one is a vertical line).

We can see that we are given the equation of the perpendicular line in slope–intercept form. The coefficient of π‘₯ is 2, so the slope of this line is 2. The slope of the line we want to find is the negative of the reciprocal of this value. We have π‘š=βˆ’12.

We know that the equation of a line with a slope of π‘š that passes through (π‘₯,𝑦) can be written in the form π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯).

Substituting π‘š=βˆ’12, π‘₯=3, and 𝑦=βˆ’3 into this equation gives us π‘¦βˆ’(βˆ’3)=βˆ’12(π‘₯βˆ’3)𝑦+3=βˆ’12(π‘₯βˆ’3).

We can now expand the brackets and rearrange to get 𝑦+3=βˆ’12π‘₯+32𝑦=βˆ’12π‘₯βˆ’32.

In our next example, we will find the equation of a line given a point on the line and two points on a perpendicular line.

Example 5: Finding the Equation of a Line given a Point on the Line and Two Points on Another Perpendicular Line

Find the equation of the straight line passing through the point (βˆ’1,1) and perpendicular to the straight line passing through the points (βˆ’9,9) and (6,βˆ’3).

Answer

We want to determine the equation of a straight line given a point on the line and two points on a perpendicular line. We can do this by recalling that the equation of a line of slope π‘š that passes through the point (π‘₯,𝑦) is π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯). We already know that our line passes through the point (βˆ’1,1), so we just need its slope.

We can find the slope of the line by recalling that its product with the slope of the line perpendicular to it will be βˆ’1. We can determine the slope of a line passing through (π‘₯,𝑦) and (π‘₯,𝑦) using the formula π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯. Substituting π‘₯=βˆ’9, 𝑦=9, π‘₯=6, and 𝑦=βˆ’3 into this formula gives us the slope π‘šοŒ of the perpendicular line: π‘š=9βˆ’(βˆ’3)(βˆ’9)βˆ’6=βˆ’45.

Taking the negative of the reciprocal of this value gives us the slope π‘š of our line: π‘š=βˆ’1π‘š=54.

We can now substitute π‘š=54, π‘₯=βˆ’1, and 𝑦=1 into the equation of a line to get π‘¦βˆ’1=54(π‘₯βˆ’(βˆ’1)).

We can now expand the brackets and rearrange to obtain π‘¦βˆ’1=54π‘₯+54𝑦=54π‘₯+54+1𝑦=54π‘₯+94.

In our next example, we will determine whether the lines between two pairs of points are parallel, perpendicular, or neither.

Example 6: Determining Whether the Lines Between Given Points Are Parallel, Perpendicular, or Neither

Given that the coordinates of the points 𝐴, 𝐡, 𝐢, and 𝐷 are (βˆ’15,8), (βˆ’6,10), (βˆ’8,βˆ’7), and (βˆ’6,βˆ’16), respectively, determine whether ⃖⃗𝐴𝐡 and ⃖⃗𝐢𝐷 are parallel, perpendicular, or neither.

Answer

We can check the relationship between a pair of lines by comparing the slopes. We recall that parallel lines have the same slopes and perpendicular lines have slopes that multiply to give βˆ’1, provided that neither line is vertical.

We can calculate the slope of a line passing through (π‘₯,𝑦) and (π‘₯,𝑦) using the formula π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯. Substituting π‘₯=βˆ’15, 𝑦=8, π‘₯=βˆ’6, and 𝑦=10 into the formula gives π‘š=8βˆ’10βˆ’15βˆ’(βˆ’6)=29.

Substituting π‘₯=βˆ’8, 𝑦=βˆ’7, π‘₯=βˆ’6, and 𝑦=βˆ’16 into the formula gives π‘š=βˆ’7βˆ’(βˆ’16)βˆ’8βˆ’(βˆ’6)=βˆ’92.

We see that the product of the slopes of the lines is βˆ’1: π‘šΓ—π‘š=29Γ—ο€Όβˆ’92=βˆ’1.

Thus, the lines are perpendicular.

In our final example, we will use the fact that adjacent sides in a rectangle are perpendicular and opposite sides are parallel to determine the coordinates of the final vertex in a rectangle given the coordinates of the other three vertices.

Example 7: Determining Whether the Lines Between Given Points Are Parallel, Perpendicular, or Neither

The vertices of a rectangle 𝐴𝐡𝐢𝐷 have the coordinates 𝐴(βˆ’3,5), 𝐡(3,7), 𝐢(6,βˆ’2), and 𝐷(π‘₯,𝑦). Determine the values of π‘₯ and 𝑦.

Answer

We first recall that the adjacent sides in a rectangle are perpendicular. This means that ⃖⃗𝐴𝐡 is perpendicular to ⃖⃗𝐢𝐷. We can find expressions for the slopes of these lines and then use the fact that the lines are parallel to find an equation involving π‘₯ and 𝑦.

We can calculate the slope of a line passing through (π‘₯,𝑦) and (π‘₯,𝑦) using the formula π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯. Substituting π‘₯=βˆ’3, 𝑦=5, π‘₯=3, and 𝑦=7 into the formula gives π‘š=5βˆ’7βˆ’3βˆ’3=13.

Substituting π‘₯=6, 𝑦=βˆ’2, π‘₯=π‘₯, and 𝑦=π‘¦οŠ¨ into the formula gives π‘š=βˆ’2βˆ’π‘¦6βˆ’π‘₯.

Since the lines are parallel, they will have the same slopes. Hence, π‘š=π‘š13=βˆ’2βˆ’π‘¦6βˆ’π‘₯.

We can rearrange this equation to get 3(βˆ’2βˆ’π‘¦)=6βˆ’π‘₯βˆ’3π‘¦βˆ’6=6βˆ’π‘₯βˆ’3𝑦=12βˆ’π‘₯.

This is not enough information to find the values of π‘₯ and 𝑦. We can follow this process again with ⃖⃗𝐴𝐡 and ⃖⃗𝐴𝐷. This time, the sides are adjacent, so they must be perpendicular.

Substituting π‘₯=βˆ’3, 𝑦=5, π‘₯=π‘₯, and 𝑦=π‘¦οŠ¨ into the formula gives π‘š=5βˆ’π‘¦βˆ’3βˆ’π‘₯.

We can note that ⃖⃗𝐴𝐡 is neither horizontal nor vertical, so ⃖⃗𝐴𝐷 will not be horizontal or vertical. Thus, the product of the slopes of the lines will be βˆ’1, since they are perpendicular: π‘šΓ—π‘š=βˆ’113Γ—5βˆ’π‘¦βˆ’3βˆ’π‘₯=βˆ’1.

We can rearrange this equation to get 5βˆ’π‘¦=βˆ’3(βˆ’3βˆ’π‘₯)5βˆ’π‘¦=3π‘₯+9βˆ’π‘¦=3π‘₯+4.

We now have a pair of simultaneous equations involving π‘₯ and 𝑦. We can solve these to determine the values of π‘₯ and 𝑦. Using the second equation, we have 𝑦=βˆ’(3π‘₯+4)=βˆ’3π‘₯βˆ’4.

We can substitute this expression for 𝑦 into the equation βˆ’3𝑦=12βˆ’π‘₯ to get βˆ’3(βˆ’3π‘₯βˆ’4)=12βˆ’π‘₯.

Expanding the brackets yields 9π‘₯+12=12βˆ’π‘₯.

We can then solve for π‘₯9π‘₯+π‘₯=12βˆ’1210π‘₯=0π‘₯=0.

Substituting π‘₯=0 into the equation 𝑦=βˆ’3π‘₯βˆ’4 gives us 𝑦=βˆ’3(0)βˆ’4=βˆ’4.

Hence, π‘₯=0 and 𝑦=βˆ’4.

Let us finish by recapping some of the important points from this explainer.

Key Points

  • If two nonvertical lines are parallel, then they have the same slopes.
  • If two distinct lines have the same slopes (π‘š=π‘š) or are both vertical, then they are parallel.
  • If two nonvertical lines are perpendicular, then their slopes are the negatives of the reciprocals of each other. Alternatively, the product of their slopes gives βˆ’1.
  • If two lines have slopes π‘šοŠ§ and π‘šοŠ¨ such that π‘š=βˆ’1π‘šοŠ§οŠ¨, then they are perpendicular.
  • Distinct vertical lines are parallel to each other.
  • Horizontal and vertical lines are perpendicular to each other.
  • We can check if the sides of polygons are parallel or perpendicular from the coordinates of their vertices by comparing the slopes of the sides.

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