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

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

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Video Transcript

In this video, we’ll learn how to find the measure of an acute angle between two straight lines in the coordinate plane. To do this, we use a formula involving the tangent of the angle between the two lines and the slopes of the two lines. This means that depending on the form of the line, we may have to work out their slopes. So let’s first remind ourselves of some of the more familiar forms of a straight line. We call it the general form of a straight line in the coordinate plane. It’s π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐 is equal to zero where π‘Ž, 𝑏, and 𝑐 are real numbers. Alternatively, when we know the slope π‘š and its 𝑦-intercept 𝑏, we can write the equation in slope–intercept form. That’s 𝑦 is equal to π‘šπ‘₯ plus 𝑏. Note that in the slope–intercept form, the constant 𝑏 is the 𝑦-intercept, and this is not the same as the 𝑏 in the general form.

Given the slope π‘š of a line and the point π‘₯ zero, 𝑦 zero on the line, we may also write the equation of the line in point–slope form. That’s 𝑦 minus 𝑦 zero is equal to π‘š times π‘₯ minus π‘₯ zero where π‘₯ zero, 𝑦 zero are the coordinates of the point on the line. Now suppose we have a line whose slope π‘š is positive. The angle πœƒ measured counterclockwise from the positive π‘₯-direction is acute. That’s between zero and 90 degrees. We know from the point–slope form that if we have two points on the line 𝑃 and 𝑄, with coordinates π‘₯ zero, 𝑦 zero and π‘₯ one, 𝑦 one, respectively, then the slope π‘š of the line is the ratio of the difference in 𝑦 to the difference in π‘₯.

And now, if we form a right-angle triangle with our two points 𝑃, 𝑄, and a third point 𝑅 in the plane, we have 𝑦 one minus 𝑦 zero is 𝑄𝑅, π‘₯ one minus π‘₯ zero is 𝑃𝑅, and with respect to our angle πœƒ, 𝑄𝑅 is the opposite side and 𝑃𝑅 is the adjacent side so that 𝑄𝑅 over 𝑃𝑅 is the tan of the angle πœƒ. Our gradient π‘š is therefore tan πœƒ. A similar argument tells us that if the angle measured counterclockwise from the positive direction of the π‘₯-axis to the line is obtuse, that is, πœƒ is between 90 degrees in 180 degrees, then the slope of the line passing through the points 𝑃 and 𝑄 is π‘š, which is equal to negative 𝑃𝑅 over 𝑄𝑅. And that’s negative tan 𝛼. And since negative the tan of 𝛼 is equal to the tan of πœƒ, we again have π‘š is equal to tan πœƒ.

This means that whether our angle is acute or obtuse when it’s measured counterclockwise from the positive π‘₯-direction, the slope of this line π‘š is equal to the tangent of the angle. It’s worth mentioning that although we can extend our method to the special cases of vertical and horizontal lines, we’re not going to cover these in this video. We simply note that horizontal lines have a slope π‘š equal to zero and an angle of zero degrees, and vertical lines have an undefined slope and an angle of 90 degrees.

Now suppose we have two lines in the coordinate plane with slopes π‘š one is equal to the tan of angle πœƒ one and π‘š two is the tan of angle πœƒ two. Well, we know that in the example shown, πœƒ one is greater than πœƒ two and both angles are acute. And assuming that the lines are not parallel, that’s π‘š one does not equal π‘š two, then since the angles in a triangle must sum to 180 degrees, we have πœƒ two plus 𝛼 plus 180 minus πœƒ one is 180. Now changing the subject of the equation to 𝛼, 180 minus 180 is equal to zero gives us 𝛼 is equal to πœƒ one minus πœƒ two. And taking the tangent on both sides, we have tan 𝛼 is equal to tan πœƒ one minus πœƒ two.

We can use the addition formula for the tangent to give us tan 𝛼 is equal to tan πœƒ one minus tan πœƒ two over one plus tan πœƒ one tan πœƒ two. And this is true for any two lines as described in the diagram. Depending on the position of the lines and the location of their point of intersection, the proof differs slightly, but now we can marry our slopes π‘š one and π‘š two with our angles πœƒ one and πœƒ two, giving us that the angle 𝛼 between two nonparallel lines in the coordinate plane with slopes π‘š one and π‘š two, such that π‘š one π‘š two is not equal to negative one, is given by tan 𝛼 is π‘š one minus π‘š two divided by one plus π‘š one π‘š two.

Now recalling that there are two angles when two lines cross, one obtuse and one acute, we refer to the smaller acute angle as the angle. And remember that the negative tangent corresponds to the larger obtuse angle. So to ensure that our tangent is the tangent of the acute angle, the smaller angle, we take the absolute value of our right-hand side. So the tan 𝛼 is the absolute value of π‘š one minus π‘š two over one plus π‘š one multiplied by π‘š two. Now let’s see how this works in an example where we’re given the slopes of two lines.

Determine to the nearest second the measure of the angle between two straight lines having slopes of five and one-quarter.

Knowing the slopes of our two lines, that’s π‘š one is equal to five and π‘š two is one-quarter, we can find the acute angle 𝛼 between the lines using the formula tan 𝛼 or the tangent of 𝛼 is the absolute value of π‘š one minus π‘š two over one plus π‘š one multiplied by π‘š two. In our case, this gives us the tan of 𝛼 is the absolute value of five minus one over four all divided by one plus five times one over four. The right-hand side evaluates to 19 divided by nine. So this is the tan of our angle 𝛼. And now taking the inverse tangent on both sides, we have 𝛼 is equal to the inverse tan of 19 over nine. And from our calculators, we find to four decimal places that 𝛼 is 64.6538 degrees.

We’re asked to find 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 begin therefore by multiplying the decimal part of our degrees by 60, and this gives us to four decimal places 39.2294 minutes. Now, multiplying the decimal part of our minutes by 60 to four decimal places, this gives us 13.7666 seconds, which is approximately 14 seconds. To the nearest second then, the angle between our two straight lines is 64 degrees, 39 minutes, and 14 seconds.

In this example, we were given the slopes of the two lines, and in our next example, we’ll see how to find the angle between two lines in the coordinate plane where the lines are both given in general form.

Find the measure of the acute angle between the two straight lines whose equations are 11π‘₯ plus 10𝑦 minus 28 is equal to zero and two π‘₯ plus 𝑦 plus 15 is equal to zero to the nearest second.

We’re given the equations for two straight lines in general form. That’s π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐 is equal to zero, where π‘Ž, 𝑏, and 𝑐 are real numbers. And to find the acute angle 𝛼 between the two lines, we’ll use the formula tan 𝛼 is equal to the absolute value of π‘š one minus π‘š two over one plus π‘š one multiplied by π‘š two, where π‘š one and π‘š two are the slopes of our two lines. And for a line in general form, we know that the slope is given by π‘š equal to negative π‘Ž over 𝑏. Our lines are given by 𝐿 one, which is 11π‘₯ plus 10𝑦 minus 28 is equal to zero, and 𝐿 two, which is two π‘₯ plus 𝑦 plus 15 is equal to zero. This means that for our line 𝐿 one, π‘Ž is equal to 11, 𝑏 is equal to 10, and 𝑐 is negative 28. This means that our slope π‘š one which is negative π‘Ž over 𝑏 is negative 11 over 10.

And now applying the same to our line 𝐿 two, where in this case π‘Ž is equal to two, 𝑏 is equal to positive one, and 𝑐 is 15, so that our slope π‘š two is equal to negative two. And now we can substitute these two into the formula for the tangent of the angle between the two lines. Now simplifying our numerator and denominator, we have the absolute value of negative 11 over 10 plus two divided by one plus 22 over 10, and this evaluates to nine divided by 32. And now making some room, we can take the inverse tangent on both sides to find the angle 𝛼; that is, 𝛼 is the inverse tan of nine over 32. And to four decimal places in our calculators, this is 15.7086 degrees.

We’re asked for the measure of the angle to the nearest second. And to find this, we recall that there are 60 minutes in one degree and 60 seconds in one minute. If we multiply the decimal part of our degrees by 60, we get 42.5182 to four decimal places, and that’s minutes. And now, multiplying the decimal part of our minutes by 60 gives us 31.0961 seconds to four decimal places. That’s approximately 31 seconds. The measure of the acute angle between the two lines to the nearest second is therefore 15 degrees, 42 minutes, and 31 seconds.

In this example, we were given two lines in general form, and before our next example, let’s remind ourselves of some other ways in which straight lines can be expressed. If we have a line passing through the point 𝐴 with coordinates π‘Ž one, π‘Ž two in the direction of the vector 𝐝 with components 𝑑 one, 𝑑 two, then in the vector form of the straight line, each unique value of the real parameter 𝑑 gives the position vector 𝐫 of a point on the line. The line in parametric form is given by π‘₯ is π‘Ž one plus 𝑑𝑑 one and 𝑦 is π‘Ž two plus 𝑑𝑑 two. And in Cartesian form, π‘₯ minus π‘Ž one over 𝑑 one is equal to 𝑦 minus π‘Ž two over 𝑑 two, where 𝑑 one and 𝑑 two are nonzero.

Note that by solving each of the parametric equations for 𝑑 and equating, we obtain the Cartesian form. And this can be rearranged to give 𝑦 is equal to 𝑑 two over 𝑑 one π‘₯ plus π‘Ž two minus 𝑑 two over 𝑑 one times π‘Ž one. And this is now in slope–intercept form, where our slope π‘š is 𝑑 two over 𝑑 one. This means that given a line in any of the forms shown and in particular its direction vector, we’re able to find its slope which is 𝑑 two over 𝑑 one provided 𝑑 one is nonzero. In our next example, we’ll use this to find the angle between two straight lines whose equations are given in vector and parametric forms.

Find the measure of the acute angle between the two straight lines 𝐿 one and 𝐿 two whose equations are 𝐫 is two, seven plus 𝐾 times negative one, eight and π‘₯ is equal to three plus 12𝑑, 𝑦 is four 𝑑 minus five, respectively, in terms of degrees, minutes, and seconds to the nearest second.

To find the acute angle 𝛼 between the two lines in the coordinate plane, we’re going to use the formula the tan of 𝛼 is the absolute value of π‘š one minus π‘š two divided by one plus π‘š one multiplied by π‘š two. This is where π‘š one is the slope of the line 𝐿 one and π‘š two is the slope of the line 𝐿 two. And this, of course, means that we must find the slopes π‘š one and π‘š two. The first of our lines 𝐿 one is given in vector form. This means that any point on the line π‘₯, 𝑦 passes through the point 𝐴 with coordinates π‘Ž one, π‘Ž two in the direction of the direction vector with components 𝑑 one and 𝑑 two for a unique value of the parameter 𝑑. The slope of the line is given by 𝑑 two over 𝑑 one, where 𝑑 one is nonzero.

In our line 𝐿 one, we can see that the constant 𝐾 corresponds to the parameter 𝑑 and that our direction vector 𝐝 has components negative one and eight. This means that 𝑑 one corresponds to negative one and 𝑑 two is eight. Our slope π‘š one is therefore eight divided by negative one, which is negative eight. Our second line 𝐿 two is given in parametric form. That’s where π‘₯ is equal to π‘Ž one plus 𝑑 times 𝑑 one and 𝑦 is π‘Ž two plus 𝑑 times 𝑑 two. And again our direction vector is 𝐝 with components 𝑑 one, 𝑑 two, and our line passes through the point 𝐴 with coordinates π‘Ž one, π‘Ž two. As before, our slope is given by 𝑑 two over 𝑑 one.

Comparing our line 𝐿 two with the parametric form, our constant 𝑑 corresponds to the parameter 𝑑 so that our direction vector has components 12, four. Our slope π‘š two, which again is 𝑑 two over 𝑑 one, is four over 12, that is, one over three. We can now use our two slopes, π‘š one is negative eight and π‘š two is one over three, to find the tan of our angle 𝛼. This evaluates to negative 25 over three divided by negative five over three. And this reduces to five. And now taking the inverse tangent on both sides, we have 𝛼 is the inverse tan of five which is approximately 78.6900 degrees.

We’re asked for the angle to the nearest second. And to find this, we recall that there are 60 minutes in one degree and 60 seconds in one minute. To find the number of minutes, we multiply the decimal part of our result by 60 which, to four decimal places, gives us 41.4040 minutes. And to find the number of seconds, we multiply the decimal part of our minutes by 60 which, to four decimal places, is 24.2430 seconds. That’s approximately 24 seconds. And making a little room, to the nearest second, the acute angle between the two straight lines 𝐿 one and 𝐿 two is 78 degrees, 41 minutes, and 24 seconds.

Let’s complete this video by reminding ourselves of some of the key points we’ve covered. Given two lines in the coordinate plane with slopes π‘š one and π‘š two, to find the acute angle 𝛼 between the two lines, we use the formula tan 𝛼 or the tangent of 𝛼 is the absolute value of π‘š one minus π‘š two divided by one plus π‘š one multiplied by π‘š two. If π‘š one multiplied by π‘š two is negative one, then the expression for tan 𝛼 is undefined because the denominator is equal to zero. This means that the lines are perpendicular, so the angle between them is 90 degrees. We refer to the acute angle between two lines in the plane as the angle between them. And finally, if the lines are parallel, they don’t intersect and so there is no angle between them.

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