Lesson Explainer: Angle between Two Straight Lines in the Coordinate Plane Mathematics

In this explainer, we will learn how to find the measure of an acute angle between two straight lines in the coordinate plane.

We recall that the equation of any straight line in the coordinate plane can be written in the form π‘Žπ‘₯+𝑏𝑦+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are constants. This is called the general form of a straight line. Alternatively, when we know the slope or gradient of a line, π‘š, and its 𝑦-intercept, 𝑏, we can write the equation in slope–intercept form: 𝑦=π‘šπ‘₯+𝑏.

(Note that the constant, 𝑏, in the slope–intercept form and the general form is not the same.)

Given the slope, π‘š, of a line and a point, (π‘₯,𝑦), on the line, we may also write the equation of the line in point–slope form: π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯).

If we have the slope, π‘š, of a line, we know that when π‘š is positive, the angle, πœƒ, measured counterclockwise from the positive direction of the π‘₯-axis to the line is acute. That is, 0<πœƒ<90∘∘. We also know from the point–slope form that if we have two distinct points on a line, say 𝑃(π‘₯,𝑦) and 𝑄(π‘₯,𝑦), then the slope of the line is the ratio of the difference in 𝑦 to the difference in π‘₯: π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.

Now, from a geometrical perspective, if we form a right triangle using our two points 𝑃 and 𝑄 and a third point 𝑅 in the plane as shown in the diagram, then recalling that tanoppositeadjacentπœƒ=, we have tanπœƒ=𝑄𝑅𝑃𝑅.

Since 𝑄𝑅 is the difference in the 𝑦-coordinates of two points on the line and 𝑃𝑅 is the difference in the corresponding π‘₯-coordinates of two points on the line, this means that tanπœƒ=π‘š.

Similarly, if the angle measured counterclockwise from the positive direction of the π‘₯-axis to the line is obtuse, that is, 90<πœƒ<180∘∘, as shown in the diagram below, then the slope of the line passing through points 𝑃 and 𝑄 is π‘š=βˆ’π‘ƒπ‘…π‘„π‘…=βˆ’π›Ό=πœƒ.tantan

Hence, whether the angle, πœƒ, measured counterclockwise from the positive direction of the π‘₯-axis to the line is acute or obtuse, the slope of the line, π‘š, is equal to tanπœƒ.

It is worth mentioning that although we can extend our method to the special cases of vertical and horizontal lines, we do not cover these in this explainer. We note simply that horizontal lines have a slope π‘š=0 and an angle of 0∘, and vertical lines have an undefined slope and an angle of 90∘.

Now, suppose we have two lines in the coordinate plane, with slopes π‘š=πœƒοŠ§οŠ§tan and π‘š=πœƒοŠ¨οŠ¨tan, for example, as shown in the diagram below. In this case, πœƒ>πœƒοŠ§οŠ¨, and both angles are acute.

Assuming that the lines are not parallel, that is, π‘šβ‰ π‘šοŠ§οŠ¨, and given that the angles in a triangle must sum to 180∘, we have πœƒ+𝛼+(180βˆ’πœƒ)=180𝛼=180βˆ’πœƒβˆ’(180βˆ’πœƒ)=πœƒβˆ’πœƒ.

Taking the tangent on both sides then gives us tantan𝛼=(πœƒβˆ’πœƒ). Now, using the trigonometric identity, tantantantantan(𝐴±𝐡)=𝐴±𝐡1βˆ“π΄π΅, we have tantantantantantan𝛼=(πœƒβˆ’πœƒ)=πœƒβˆ’πœƒ1+πœƒπœƒ.

This is true for any two lines described as above. However, while the proof differs slightly depending on the position of the lines and the location of their point of intersection, provided neither line is vertical, that is, neither πœƒοŠ§ nor πœƒ=90∘, this result holds for any two nonparallel lines with angles πœƒοŠ§ and πœƒοŠ¨, measured counterclockwise from the positive direction of the π‘₯-axis, to each line, respectively.

Hence, since π‘š=πœƒοŠ§οŠ§tan and π‘š=πœƒοŠ¨οŠ¨tan, this leads us to the following definition.

Definition: The Angle between Two Straight Lines in the Coordinate Plane

An angle 𝛼 between two nonparallel lines in the coordinate plane with slopes π‘šοŠ§ and π‘šοŠ¨, such that π‘šπ‘šβ‰ βˆ’1, is given by tan𝛼=π‘šβˆ’π‘š1+π‘šπ‘š.

If the lines are parallel, then π‘š=π‘šοŠ§οŠ¨ and there is no angle between them. If the lines are neither parallel nor perpendicular, then there are two angles between them. We refer to the smaller angle as β€œthe angle” or β€œthe acute angle” between the lines.

A negative tangent corresponds to the larger, obtuse angle, π›ΌοŠ¨, and to ensure that the tangent is that of the acute angle, π›ΌοŠ§, we take the absolute value. Hence, the tangent of the acute angle between two lines in the coordinate plane is tan𝛼=|||π‘šβˆ’π‘š1+π‘šπ‘š|||.

Note that if π‘šπ‘š=βˆ’1, then the denominator is equal to 0, and this expression is undefined. In this case, the lines are perpendicular, and 𝛼=90∘.

Let’s see how this works in an example where we are given the slopes of two lines.

Example 1: Finding the Measure of the Angle between Two Straight Lines given Their Slopes

Determine, to the nearest second, the measure of the acute angle between two straight lines having slopes of 5 and 14.

Answer

Knowing the slopes, π‘šοŠ§ and π‘šοŠ¨, of two lines in the coordinate plane, we can find the acute angle, 𝛼, between the lines using the formula tan𝛼=|||π‘šβˆ’π‘š1+π‘šπ‘š|||.

Letting π‘š=5 and π‘š=14, we have tan𝛼=||||5βˆ’1+5β‹…||||.οŠͺοŠͺ

Evaluating the right-hand side then leads to tan𝛼=199, and taking the inverse tangent on both sides gives us 𝛼=ο€Ό199=64.6538….tan∘

We are asked to find the measure of the angle to the nearest second, and to do this, we recall that there are 60 minutes in one degree and 60 seconds in one minute. We therefore multiply the decimal part of our degrees by 60: 0.6538…×60=39.2294…. Hence, we have 39.2294…′ (minutes), and multiplying the decimal part of our minutes by 60 gives us 0.2294…×60=13.7666β€¦β‰ˆ14β€²β€² (seconds).

The measure of the acute angle between the two lines, to the nearest second, is therefore 6439β€²14β€²β€²βˆ˜.

Our next example demonstrates how to find the angle between two lines in the coordinate plane, where the lines are both given in general form.

Example 2: Finding the Angle between Two Straight Lines in Two Dimensions

Find the measure of the acute angle between the two straight lines whose equations are 11π‘₯+10π‘¦βˆ’28=0 and 2π‘₯+𝑦+15=0 to the nearest second.

Answer

We are given two lines whose equations are in general form, that is, in the form π‘Žπ‘₯+𝑏𝑦+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are real numbers. To find the acute angle, 𝛼, between the two lines, we will use the formula tan𝛼=|||π‘šβˆ’π‘š1+π‘šπ‘š|||, where π‘šοŠ§ and π‘šοŠ¨ are the slopes of the given lines. We will therefore, need to find these two slopes given in each case by βˆ’π‘Žπ‘.

Our lines are 𝐿∢11π‘₯+10π‘¦βˆ’28=0,𝐿∢2π‘₯+𝑦+15=0, whose slopes are therefore π‘š=βˆ’1110 and π‘š=βˆ’2.

Substituting these two values into the formula for tan𝛼, we then have tan𝛼=||||βˆ’(βˆ’2)1+(βˆ’2)||||=||||+21+||||=932.

Now, taking the inverse tangent of both sides, we find 𝛼=ο€Ό932=15.7086….tan∘

We are asked for the measure of the angle to the nearest second. To find this, we recall that there are 60 minutes in one degree and 60 seconds in one minute. We therefore multiply the decimal part of our degrees by 60: 0.7086…×60=42.5182…. Hence, we have 42.5182…′ (minutes), and multiplying the decimal part of our minutes by 60 gives us 0.5182…×60=31.0961β€¦β‰ˆ31β€²β€² (seconds).

The measure of the acute angle between the two lines, to the nearest second, is therefore, 1542β€²31β€²β€²βˆ˜.

In our next example, we find the acute angle between one line whose equation is given in general form and a second line passing through two known points.

Example 3: Finding the Angle between Two Straight Lines in Two Dimensions

Determine the measure of the acute angle between the straight line π‘₯βˆ’π‘¦+4=0 and the straight line passing through the points (3,βˆ’2) and (βˆ’2,4) to the nearest second.

Answer

To find the acute angle, 𝛼, between the two given lines, we can use the formula tan𝛼=|||π‘šβˆ’π‘š1+π‘šπ‘š|||, where π‘šοŠ§ is the slope of our first line, which we will denote 𝐿, and π‘šοŠ¨ is the slope of the second line, which we will denote 𝐿. To use this formula, we will need to find the slopes of the two lines.

Our first line is given in the general form: π‘Žπ‘₯+𝑏𝑦+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are real numbers and where, in our case, π‘Ž=1, 𝑏=βˆ’1, and 𝑐=4. The slope, π‘šοŠ§, is given by βˆ’π‘Žπ‘; hence, π‘š=1.

To find our second slope, π‘šοŠ¨, we recall that, given two distinct points on a line, (π‘₯,𝑦) and (π‘₯,𝑦), the slope of the line through the points is given by the change in 𝑦 divided by the change in π‘₯. That is, π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.

For our second line, 𝐿, we are given the coordinates of two points on the line, (3,βˆ’2) and (βˆ’2,4). The slope, π‘šοŠ¨, of 𝐿 is therefore π‘š=4βˆ’(βˆ’2)βˆ’2βˆ’3=βˆ’65.

Now that we have the two slopes π‘š=1 and π‘š=βˆ’65, we can use the stated formula for the tangent of the acute angle, 𝛼, between the two lines: tan𝛼=||||1βˆ’ο€»ο‡1+1×||||=||||1+1βˆ’||||=|βˆ’11|=11.

Now, taking the inverse tangent, we find 𝛼=(11)=84.8055….tan∘

We are asked for the angle to the nearest second. To find this, we recall that there are 60 minutes in one degree and 60 seconds in one minute. We therefore multiply the decimal part of our degrees by 60: 0.8055…×60=48.3342…. Hence, we have 48.3342…′ (minutes), and multiplying the decimal part of our minutes by 60 gives us 0.3342…×60=20.0559β€¦β‰ˆ20β€²β€² (seconds).

The measure of the angle between the two lines, to the nearest second, is therefore 8448β€²20β€²β€²βˆ˜.

Up till now, our lines have been defined either in general form or via known points on the line from which we were able to find the slopes. However, there are other forms in which straight lines can be expressed, as in the following definition.

Definition: The Vector, Parametric, and Cartesian Forms of a Line in the Coordinate Plane

A line passing through a point A with position vector βƒ‘π‘Ž=(π‘Ž,π‘Ž), in the direction of the vector ⃑𝑑=(𝑑,𝑑), can be written in the following forms: Vectorform:Parametricform:Cartesianform:βƒ‘π‘Ÿ=βƒ‘π‘Ž+𝑑⃑𝑑(π‘₯,𝑦)=(π‘Ž,π‘Ž)+𝑑(𝑑,𝑑),π‘₯=π‘Ž+𝑑𝑑𝑦=π‘Ž+𝑑𝑑,π‘₯βˆ’π‘Žπ‘‘=π‘¦βˆ’π‘Žπ‘‘.

In the vector form, each unique value of the real parameter 𝑑 gives the position vector βƒ‘π‘Ÿ=(π‘₯,𝑦) of a point on the line, and in the Cartesian form, we assume both π‘‘οŠ§ and π‘‘οŠ¨ are nonzero.

Note that by solving each of the parametric equations for 𝑑 and equating, we obtain the Cartesian form, which can be rearranged as follows: π‘₯βˆ’π‘Žπ‘‘=π‘¦βˆ’π‘Žπ‘‘π‘¦=𝑑𝑑(π‘₯βˆ’π‘Ž)+π‘Ž=𝑑𝑑π‘₯+ο€½π‘Žβˆ’π‘‘π‘‘π‘Žο‰.

This is now in slope–intercept form, 𝑦=π‘šπ‘₯+𝑏, where here the slope π‘š=π‘‘π‘‘οŠ¨οŠ§. Hence, given a line in any of the above forms and, in particular, its direction vector, we are able to find its slope, which is π‘‘π‘‘οŠ¨οŠ§, provided π‘‘οŠ§ is nonzero.

In our next example, we will use this to find the angle between two lines whose equations are given in vector and parametric forms.

Example 4: Finding the Angle between Two Straight Lines in Two Dimensions

Find the measure of the acute angle between the two straight lines 𝐿 and 𝐿 whose equations are βƒ‘π‘Ÿ=(2,7)+𝐾(βˆ’1,8) and π‘₯=3+12𝑑, 𝑦=4π‘‘βˆ’5, respectively, in terms of degrees, minutes, and seconds, to the nearest second.

Answer

To find the acute angle, 𝛼, between two lines in the coordinate plane, we use the formula tan𝛼=|||π‘šβˆ’π‘š1+π‘šπ‘š|||, where π‘šοŠ§ and π‘šοŠ¨ are the slopes of the two lines. We must therefore find the slopes of the two given lines.

The first of our two lines, πΏβˆΆβƒ‘π‘Ÿ=(2,7)+𝐾(βˆ’1,8), is given in vector form, that is, in the form βƒ‘π‘Ÿ=(π‘₯,𝑦)=(π‘Ž,π‘Ž)+𝑑(𝑑,𝑑), where the line passes through the point with position vector βƒ‘π‘Ž=(π‘Ž,π‘Ž), in the direction of the vector ⃑𝑑=(𝑑,𝑑). The slope, π‘š, of a line with direction vector (𝑑,𝑑) is π‘‘π‘‘οŠ¨οŠ§.

For our line, 𝐿, we see that the constant 𝐾 corresponds to the parameter 𝑑 so that our direction vector is ⃑𝑑=(βˆ’1,8). The slope, π‘šοŠ§, of this line is therefore π‘š=𝑑𝑑=8βˆ’1=βˆ’8.

Our second line, 𝐿, π‘₯=3+12𝑑,𝑦=4π‘‘βˆ’5, is given in parametric form. That is, in the form π‘₯=π‘Ž+𝑑𝑑,𝑦=π‘Ž+𝑑𝑑, where again the line passes through the point with position vector βƒ‘π‘Ž=(π‘Ž,π‘Ž), in the direction of the vector ⃑𝑑=(𝑑,𝑑). Comparing, we see that parameter 𝑑 in 𝐿 corresponds to parameter 𝑑 in the general parametric equations and that our direction vector is therefore ⃑𝑑=(12,4). Hence, the slope of line 𝐿 is given by π‘š=𝑑𝑑=412=13.

We can now use our two slopes, π‘š=βˆ’8 and π‘š=13, in the formula to find the acute angle between the two lines: tan𝛼=|||π‘šβˆ’π‘š1+π‘šπ‘š|||=||||βˆ’8βˆ’1+(βˆ’8)||||=5.

Now, taking the inverse tangent on both sides, we have 𝛼=(5)=78.6900….tan∘

We are asked for the angle to the nearest second. To find this we recall that there are 60 minutes in one degree and 60 seconds in one minute. We therefore multiply the decimal part of our degrees by 60: 0.6900…×60=41.4040…. Hence, we have 41.4040…′ (minutes), and multiplying the decimal part of our minutes by 60 gives us 0.4040…×60=24.2430β€¦β‰ˆ24β€²β€² (seconds).

Therefore, the measure of the acute angle between the two lines 𝐿 and 𝐿, to the nearest second, is 7841β€²24β€²β€²βˆ˜.

We will use the formula for the tangent of the angle between two lines, in our next example, to find the equations of the lines.

Example 5: Finding the Equations of Two Straight Lines in Two Dimensions Using the Tangent of the Angle between Them

Let πœƒ be the angle between two lines that pass through (4,βˆ’2). If tanπœƒ=121 and the slopes of the lines are π‘š and 45π‘š, with π‘š>0, find the equations of these lines.

Answer

Recalling that by the angle between two lines, we mean the smaller of the two angles, we are told that two lines, say 𝐿 and 𝐿, both pass through the point (4,βˆ’2) and that the tangent of the angle, πœƒ, between them is equal to 121. We are also given that the slope of 𝐿 is equal to π‘š and that the slope of 𝐿 is equal to 45π‘š, with π‘š>0.

To find the equations of lines 𝐿 and 𝐿, we first use the information we have in the formula for the tangent of the angle between two lines to find any possible values of π‘š. We can then use the given point and the point–slope form of a line to find the equations of lines 𝐿 and 𝐿.

Recall that the formula for the tangent of the acute angle, πœƒ, between two lines is as follows: tanπœƒ=|||π‘šβˆ’π‘š1+π‘šπ‘š|||, where π‘šοŠ§ and π‘šοŠ¨ are the slopes of the two lines. Substituting our values for tanπœƒ, π‘šοŠ§, and π‘šοŠ¨, we have 121=||||π‘šβˆ’π‘š1+π‘šβ‹…π‘š||||=π‘š5+4π‘š,οŠͺοŠͺ noting that since π‘š is positive, the right-hand side is positive. Rearranging this results in the quadratic equation 4π‘šβˆ’21π‘š+5=0, which we can solve for π‘š. Using the quadratic formula or otherwise, we find that there are two solutions: π‘š=5 and π‘š=14.

The point–slope form of a line is π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯), where π‘š is the slope, and the line passes through the point (π‘₯,𝑦). Using this with each of our solutions for π‘š, together with the point (4,βˆ’2) through which both lines pass, will produce equations for 𝐿 and 𝐿.

Beginning with our solution, π‘š=5, we recall that the slope of 𝐿 is π‘š=5 and the slope of 𝐿=45Γ—5=4. We therefore have πΏβˆΆπ‘¦βˆ’(βˆ’2)=5(π‘₯βˆ’4)𝑦+2=5π‘₯βˆ’20βˆ’5π‘₯+𝑦+22=0,πΏβˆΆπ‘¦βˆ’(βˆ’2)=4(π‘₯βˆ’4)𝑦+2=4π‘₯βˆ’164π‘₯βˆ’π‘¦βˆ’18=0.

Now with the second solution, π‘š=14, we have that the slope of 𝐿 is π‘š=14 and the slope of 𝐿=45β‹…14=15.

Therefore, in this case, we have πΏβˆΆπ‘¦βˆ’(βˆ’2)=14(π‘₯βˆ’4)𝑦+2=14π‘₯βˆ’1π‘₯βˆ’4π‘¦βˆ’12=0,πΏβˆΆπ‘¦βˆ’(βˆ’2)=15(π‘₯βˆ’4)𝑦+2=15π‘₯βˆ’45π‘₯βˆ’5π‘¦βˆ’14=0.

Hence, there are two possible pairs for the equations of the lines: βˆ’5π‘₯+𝑦+22=04π‘₯βˆ’π‘¦βˆ’18=0and or π‘₯βˆ’4π‘¦βˆ’12=0π‘₯βˆ’5π‘¦βˆ’14=0.and

We complete this explainer by reminding ourselves of some of the key points.

Key Points

  • To find the acute angle, 𝛼, between two straight lines with slopes π‘šοŠ§ and π‘šοŠ¨ in the coordinate plane, we use the formula tan𝛼=|||π‘šβˆ’π‘š1+π‘šπ‘š|||.
  • If π‘šπ‘š=βˆ’1, then the denominator in the formula for tan𝛼 is equal to 0, and the expression is undefined. In this case, the lines are perpendicular, and 𝛼=90∘.
  • We refer to the acute angle between two lines in the coordinate plane as the angle between the lines.
  • If the lines are parallel, they do not intersect, so there is no angle between them.

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