### Video Transcript

Let π be the angle between two
lines that pass through four, negative two. If tan of π equals one over 21 and
the slope of the lines are π and four-fifths π, with π greater than zero, find
the equations of these lines.

Letβs recall that by the angle
between two lines we mean the smaller of the two angles. We can define the two lines as πΏ
sub one and πΏ sub two. We are given some information about
these two lines, firstly that they pass through this point four, negative two. And weβre also told that the
tangent of the angle between them is equal to one over 21. We are further given that the
slopes, or gradients, of these two lines are π and four-fifths π, where π is
greater than zero.

Before we can find the equations of
πΏ sub one and πΏ sub two, the first thing we can do is work out the value of
π. To do this, we can recall that the
formula for the tan of the acute angle π between two lines is given as tan of π
equals the absolute value of π sub one minus π sub two over one plus π sub one π
sub two, where π sub one and π sub two are the slopes of the two lines. We can therefore substitute in the
given values for the tan of π and π sub one and π sub two. This gives us one over 21 is equal
to the absolute value of π minus four-fifths π over one plus π times four-fifths
π.

We can then simplify the right-hand
side and multiply the numerator and denominator by five to give us that one over 21
is equal to π over five plus four π squared. And we can note that since π is
positive, then the right-hand side of this equation is also positive. By taking the cross product, we
have five plus four π squared is equal to 21π. Then, by further rearranging, we
can see that we have a quadratic equation in π that we can solve for π. At this point, we might realize
that we canβt factorize this equation, so we can use a method, for example, the
quadratic formula, to help us do this.

The quadratic formula states that
π₯ is equal to negative π plus or minus the square root of π squared minus four
ππ all over two π for equations of the form ππ₯ squared plus ππ₯ plus π is
equal to zero. In our equation, we have the
variable π instead of π₯. But we can see from the equation
that π is equal to four, π is equal to negative 21, and π is equal to five. So we can substitute these values
into the quadratic formula and simplify. When we do this, we produce two
different values of π. And when weβre solving this, itβs
helpful to remember that the square root of 361 is 19. Therefore, the two values of π are
five and one-quarter.

Itβs important to realize that at
this point even though we have found two different values of the slope π, this
isnβt the same as the slopes of the original lines, which we are told were π and
four-fifths π. In fact, what we will end up with
are two different pairs of lines that meet all the criteria in the question. We can now clear some space to find
the two different pairs of equations for πΏ sub one and πΏ sub two.

Given that we have a point that
these lines pass through and the slope of the line, we can use the pointβslope form
of the equation. π¦ minus π¦ sub zero equals π
times π₯ minus π₯ sub zero, where π is the slope and the line passes through π₯ sub
zero, π¦ sub zero. Notice that we will have to be a
little bit careful when weβre using this pointβslope form that we donβt get confused
with this general letter π in the formula and the variable π and four-fifths π
that we were using in the question. Letβs take the instance of π which
is just equal to five. So we know that one of our lines
has a slope which is just equal to π, which is five.

Given that this line passes through
the point four, negative two, when we use the pointβslope form, weβll have π¦ minus
negative two is equal to five times π₯ minus four. Simplifying this and then
collecting the like terms, we have that the first line is negative five π₯ plus π¦
plus 22 is equal to zero. We still need to find the second
line when π is equal to five. But this time, we know that the
slope is actually equal to four-fifths of π. Since π is equal to five, then the
slope of the second line will be equal to four. Itβs still going through the point
four, negative two. So this time on the right-hand
side, weβll be multiplying outside these parentheses by the value four. Simplifying, we have the equation
of the line as four π₯ minus π¦ minus 18 is equal to zero.

These two lines form one possible
set of lines, which meet all the given criteria in the question. However, we also know that these
criteria will still be met when the value of π is equal to one-quarter. So letβs use the pointβslope form
of the line again to work out the other pair of equations. For the equation of the first line,
we know that the slope is simply equal to π, which is one-quarter. We can then rearrange this to give
the first equation of the line, noticing that if we want to leave it in the standard
form when weβre rearranging, weβll need to multiply by four throughout the
equation. We can then do the same to find the
second equation.

Remember that as weβve seen
previously, in this line, the value of π that weβre substituting into the
pointβslope form isnβt simply one-quarter because we know that the slope of the
second line is four-fifths of π. Four-fifths of one-quarter is
one-fifth. When we substitute this value in
along with the coordinate four, negative two, we get the equation of the second
line. Therefore, we can give the answer
of the two possible pairs of equations. π₯ minus four π¦ minus 12 equals
zero and π₯ minus five π¦ minus 14 equals zero or negative five π₯ plus π¦ plus 22
equals zero and four π₯ minus π¦ minus 18 equals zero.

In order to demonstrate the
difference between these two equation pairs, we could draw a quick sketch of each
situation. Drawing the first pair of lines
would result in this figure, and the second pair of lines would result in this
figure. In both sets of graphs, the
gradient, or slope, is greater than zero as it was supposed to be. And itβs also worth noting that the
inverse tan of one over 21 actually produces a very small angle in degrees. Itβs approximately 2.7 degrees.

So by applying the formula to find
the acute angle between two straight lines and given the relationship between their
slopes, we have found two different pairs of straight lines passing through the
coordinate four, negative two.