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 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 is
equal to zero where 𝑎, 𝑏, and 𝑐 are real numbers. Alternatively, when we know the
slope 𝑚 and its 𝑦-intercept 𝑏, we can write the equation in slope–intercept
form. That’s 𝑦 is equal to 𝑚𝑥 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 𝑚 of a line and
the point 𝑥 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
𝑚 times 𝑥 minus 𝑥 zero where 𝑥 zero, 𝑦 zero are the coordinates of the point on
the line. Now suppose we have a line whose
slope 𝑚 is positive. The angle 𝜃 measured
counterclockwise from the positive 𝑥-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 𝑥 zero, 𝑦 zero
and 𝑥 one, 𝑦 one, respectively, then the slope 𝑚 of the line is the ratio of the
difference in 𝑦 to the difference in 𝑥.
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, 𝑥 one minus 𝑥 zero is 𝑃𝑅, and with respect to our
angle 𝜃, 𝑄𝑅 is the opposite side and 𝑃𝑅 is the adjacent side so that 𝑄𝑅 over
𝑃𝑅 is the tan of the angle 𝜃. Our gradient 𝑚 is therefore tan
𝜃. A similar argument tells us that if
the angle measured counterclockwise from the positive direction of the 𝑥-axis to
the line is obtuse, that is, 𝜃 is between 90 degrees in 180 degrees, then the slope
of the line passing through the points 𝑃 and 𝑄 is 𝑚, 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 𝜃, we again have 𝑚 is equal to tan 𝜃.
This means that whether our angle
is acute or obtuse when it’s measured counterclockwise from the positive
𝑥-direction, the slope of this line 𝑚 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 𝑚 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 𝑚 one is equal to the tan of angle 𝜃 one and 𝑚
two is the tan of angle 𝜃 two. Well, we know that in the example
shown, 𝜃 one is greater than 𝜃 two and both angles are acute. And assuming that the lines are not
parallel, that’s 𝑚 one does not equal 𝑚 two, then since the angles in a triangle
must sum to 180 degrees, we have 𝜃 two plus 𝛼 plus 180 minus 𝜃 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 𝜃 one minus
𝜃 two. And taking the tangent on both
sides, we have tan 𝛼 is equal to tan 𝜃 one minus 𝜃 two.
We can use the addition formula for
the tangent to give us tan 𝛼 is equal to tan 𝜃 one minus tan 𝜃 two over one plus
tan 𝜃 one tan 𝜃 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 𝑚 one and 𝑚 two with our angles 𝜃 one and 𝜃 two,
giving us that the angle 𝛼 between two nonparallel lines in the coordinate plane
with slopes 𝑚 one and 𝑚 two, such that 𝑚 one 𝑚 two is not equal to negative one,
is given by tan 𝛼 is 𝑚 one minus 𝑚 two divided by one plus 𝑚 one 𝑚 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 𝑚 one minus 𝑚 two over one plus 𝑚 one multiplied by 𝑚 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 𝑚 one is equal to five and 𝑚 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 𝑚 one minus 𝑚 two over one plus 𝑚 one multiplied by 𝑚
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 11𝑥 plus 10𝑦 minus 28 is equal
to zero and two 𝑥 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 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 is
equal to zero, where 𝑎, 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 𝑚 one minus 𝑚 two over one plus 𝑚 one multiplied by 𝑚 two, where 𝑚 one and
𝑚 two are the slopes of our two lines. And for a line in general form, we
know that the slope is given by 𝑚 equal to negative 𝑎 over 𝑏. Our lines are given by 𝐿 one,
which is 11𝑥 plus 10𝑦 minus 28 is equal to zero, and 𝐿 two, which is two 𝑥 plus
𝑦 plus 15 is equal to zero. This means that for our line 𝐿
one, 𝑎 is equal to 11, 𝑏 is equal to 10, and 𝑐 is negative 28. This means that our slope 𝑚 one
which is negative 𝑎 over 𝑏 is negative 11 over 10.
And now applying the same to our
line 𝐿 two, where in this case 𝑎 is equal to two, 𝑏 is equal to positive one, and
𝑐 is 15, so that our slope 𝑚 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 𝑎 one, 𝑎 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 𝑡 gives the position vector 𝐫 of a point on the
line. The line in parametric form is
given by 𝑥 is 𝑎 one plus 𝑡𝑑 one and 𝑦 is 𝑎 two plus 𝑡𝑑 two. And in Cartesian form, 𝑥 minus 𝑎
one over 𝑑 one is equal to 𝑦 minus 𝑎 two over 𝑑 two, where 𝑑 one and 𝑑 two are
nonzero.
Note that by solving each of the
parametric equations for 𝑡 and equating, we obtain the Cartesian form. And this can be rearranged to give
𝑦 is equal to 𝑑 two over 𝑑 one 𝑥 plus 𝑎 two minus 𝑑 two over 𝑑 one times 𝑎
one. And this is now in slope–intercept
form, where our slope 𝑚 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 𝑥 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 𝑚 one minus 𝑚 two divided by one plus 𝑚 one multiplied
by 𝑚 two. This is where 𝑚 one is the slope
of the line 𝐿 one and 𝑚 two is the slope of the line 𝐿 two. And this, of course, means that we
must find the slopes 𝑚 one and 𝑚 two. The first of our lines 𝐿 one is
given in vector form. This means that any point on the
line 𝑥, 𝑦 passes through the point 𝐴 with coordinates 𝑎 one, 𝑎 two in the
direction of the direction vector with components 𝑑 one and 𝑑 two for a unique
value of the parameter 𝑡. 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 𝑡 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 𝑚 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 𝑥 is equal to 𝑎 one
plus 𝑡 times 𝑑 one and 𝑦 is 𝑎 two plus 𝑡 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 𝑎 one, 𝑎 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 𝑡 so that our
direction vector has components 12, four. Our slope 𝑚 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, 𝑚
one is negative eight and 𝑚 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 𝑚 one and 𝑚 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 𝑚
one minus 𝑚 two divided by one plus 𝑚 one multiplied by 𝑚 two. If 𝑚 one multiplied by 𝑚 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.
