### Video Transcript

In this video, we will learn how to
find and write the equation of a straight line in general form.

There are many forms in which the
equation of a straight line can be written, and we may already be familiar with some
of them. For example, thereβs the
slopeβintercept form, π¦ equals ππ₯ plus π, where π represents the slope of the
line and π represents its π¦-intercept. Another form in which we can give
the equation of a straight line is the pointβslope form, π¦ minus π¦ one equals π
multiplied by π₯ minus π₯ one, where π represents the slope of the line as before
and π₯ one, π¦ one represents the coordinates of any point that lies on the
line.

These two forms of the equation of
a straight line are useful in different contexts. For example, from the
slopeβintercept form, we can identify both the slope and π¦-intercept of a straight
line at a glance. But these forms do also have their
limitations mainly because it isnβt possible to write the equation of every straight
line in either of these forms, in particular, vertical lines of equations of the
form π₯ equals π for some constant value of π. This can be rearranged to π₯ minus
π equals zero. But it isnβt possible to rearrange
this equation into either of the two forms weβve already discussed because it has no
π¦-term, or at least the coefficient of π¦ is zero.

Weβll therefore introduce a third
form of the equation of a straight line. This is called the general form
because the equation of every straight line can be written in this form. The general form of the equation of
a straight line then is ππ₯ plus ππ¦ plus π equals zero, where π, π, and π are
real constants which may be equal to zero. We note that all three terms are on
the same side of the equation, and the right-hand side is equal to zero.

It isnβt required that π, π, and
π are integers, but usually when we give our answers, we do ensure that they
are. In the case of vertical lines,
which we were concerned about earlier, their equations can be rearranged to π₯ minus
π equals zero. And this is in the general form
with π, the coefficient of π₯, equal to one, π, the coefficient of π¦, equal to
zero, and π, the constant term, equal to negative π.

In the same way, the equations of
horizontal lines which we can write as π¦ equals some constant π two can be
rearranged to π¦ minus π two is equal to zero. And this is also in the general
form, this time with π equal to zero, π equal to one, and π equal to negative π
two. The equation of any diagonal line
can also be written in this form. And weβll see examples of this
during the video. Now this form, the general equation
of a straight line, is very closely related to the standard form of the equation of
a straight line. This is π΄π₯ plus π΅π¦ equals πΆ,
where π΄, π΅, and πΆ are integers and πΆ is nonnegative. We must ensure that we are clear on
the distinction between the two, which is predominantly in the position of the
constant term.

One drawback of the general form of
the equation of a straight line is that we cannot directly identify the slope of a
line given in this form from its equation. In order to find the slope, we need
to rearrange the equation into the slopeβintercept form and then read off the
coefficient of π₯. Weβll demonstrate how to do this in
our first example.

A straight line has the equation
negative 15π₯ plus three π¦ minus 12 equals zero. What is the slope of this line?

The equation of this straight line
has been given in the general form. Thatβs ππ₯ plus ππ¦ plus π
equals zero with π equal to negative 15, π equal to positive three, and π equal
to negative 12. To determine the slope of this
line, we can rearrange its equation to the slopeβintercept form. Thatβs π¦ equals ππ₯ plus π,
where the value of π, the coefficient of π₯, is the slope of the line. Now the value of π in this
equation gives the π¦-intercept. And it isnβt necessarily the same
as the value of π in the general form, but these are the letters usually used, so
weβll stick with them.

So weβll take the equation weβve
been given, and weβll rearrange it to make π¦ the subject. First, we need to isolate the
π¦-term, which we can do by adding 15π₯ and 12 to each side of the equation. That gives three π¦ equals 15π₯
plus 12. We then divide the entire equation
by three, which gives π¦ equals five π₯ plus four. This equation is now in the
slopeβintercept form. And we can identify that the
coefficient of π₯ is five. So the slope of the straight line
with equation negative 15π₯ plus three π¦ minus 12 equals zero is five.

In the example we just considered,
we found the slope of a line by rearranging its equation from the general form to
the slopeβintercept form. We found that the slope of this
line was five, and this came from simplifying the fraction 15 over three. Now, that value of 15 in the
numerator is the negative of the coefficient of π₯ in the general form, and the
value of three in the denominator is the coefficient of π¦. This illustrates a general
result. If we were to rearrange the general
form of the equation of a straight line still in its algebraic form, we would obtain
π¦ equals negative π over π π₯ minus π over π. The slope of this line will be
given by the coefficient of π₯, which is negative π over π, and so we can state a
general result.

The slope of the line ππ₯ plus
ππ¦ plus π equals zero, where π is nonzero, is negative π over π. This provides a shortcut for
finding the slope of a straight line given in the general form. We can also identify that the
π¦-intercept of this line is negative π over π, and again π must be nonzero. Letβs now consider an example in
which we will determine both the π₯- and π¦-intercepts of a straight line from its
general form.

What are the π₯-intercept and
π¦-intercept of the line three π₯ plus two π¦ minus 12 equals zero?

Weβll recall first that the π₯- and
π¦-intercept of a line are the values of π₯ and π¦ at which the line intersects that
axis. The π₯-intercept has a
π¦-coordinate of zero, and the π¦-intercept has an π₯-coordinate of zero. To obtain the value of the
π₯-intercept first then, we can substitute π¦ equals zero into the equation of the
straight line. This gives three π₯ plus two
multiplied by zero minus 12 is equal to zero, or simply three π₯ minus 12 equals
zero. We can then solve this equation to
find the value of π₯. First, we add 12 to each side,
giving three π₯ equals 12. And then we divide both sides of
the equation by three, giving π₯ equals four. And so the π₯-intercept of this
straight line is four.

To find the π¦-intercept, we
substitute π₯ equals zero into the equation of the line. And this gives three multiplied by
zero plus two π¦ minus 12 equals zero. Thatβs just two π¦ minus 12 equals
zero. And then we solve for π¦. First, we add 12 to each side of
the equation and then we divide by two, giving π¦ is equal to six. And so the π¦-intercept of this
line is six. We can state then that the
π₯-intercept and π¦-intercept of the line three π₯ plus two π¦ minus 12 equals zero
are four and six, respectively.

Weβve now seen how to find the
slope and the π₯- and π¦-intercepts of a straight line from its equation given in
the general form. Letβs now consider how we can find
the equation of a straight line in the general form given its slope and its
π¦-intercept.

Write the equation of the line with
slope three over two and π¦-intercept zero, three in the form ππ₯ plus ππ¦ plus π
equals zero.

Weβre given both the slope and the
π¦-intercept of this straight line. We can therefore write down its
equation by recalling first the slopeβintercept form of the equation of a straight
line. Thatβs π¦ equals ππ₯ plus π,
where π represents the slope of the line and π represents its π¦-intercept. The slope of this straight line is
three over two, and the π¦-coordinate of its π¦-intercept is three. So thatβs the value of π. Substituting three over two for π
and three for π then, we have π¦ equals three over two π₯ plus three.

Now, this is the equation of the
line weβre looking for, but weβve been asked to give our answer in a specific form:
ππ₯ plus ππ¦ plus π equals zero. This is the general form of the
equation of a straight line. We therefore need to rearrange the
equation weβve written. We note first that in the general
term, all three terms are on the same side of the equation. So weβll begin by subtracting the
π¦-term in our equation from both sides. When we do this, we obtain zero is
equal to three over two π₯ minus π¦ plus three. And we can, of course, write the
two sides of the equation the other way around.

Now, although this isnβt strictly
necessary, itβs usual that the values of π, π, and π in the general form are
integers. We have a coefficient of π₯ which
is a fraction. So in order to eliminate the
denominator of two, we can multiply every term in our equation by two. This gives three π₯ minus two π¦
plus six is equal to zero. This is now in the general form
with π, the coefficient of π₯, equal to three, π, the coefficient of π¦, equal to
negative two, and π, the constant term, equal to positive six. So we found that the equation of
the line with slope three over two and π¦-intercept zero, three in the general form
is three π₯ minus two π¦ plus six equals zero.

Letβs now consider an example in
which we find the equation of a straight line in general form, given a different set
of information. This time, the information given
will be the coordinates of two points that lie on the line.

Find the equation of the line that
passes through the points π΄ negative 10, two and π΅ zero, five, giving your answer
in the form ππ₯ plus ππ¦ plus π equals zero.

Weβve been given the coordinates of
two points that lie on this line. The form weβve been asked to give
our answer in is the general form of the equation of a straight line: ππ₯ plus ππ¦
plus π equals zero, where π, π, and π are real constants. To begin with, though, weβll
approach this problem using the slopeβintercept form of the equation of a straight
line. Thatβs π¦ equals ππ₯ plus π,
where π is the slope of the line and π is its π¦-intercept. We should be aware that this value
of π isnβt necessarily the same as the value of π in the general form, but these
are the letters that are generally used for both.

Now we know that this line passes
through the point π΅, which has coordinates zero, five. This is the point on the
π¦-axis. And so the π¦-intercept of the line
is five. This means that the value of π in
our slopeβintercept form is also five. And so the equation is now of the
form π¦ equals ππ₯ plus five. And we need to calculate the slope
of the line.

To do this, we recall that the
slope of a line is its change in π¦ divided by its change in π₯. If we have two points which lie on
a line with coordinates π₯ one, π¦ one and π₯ two, π¦ two, then the slope can be
calculated as π¦ two minus π¦ one over π₯ two minus π₯ one. We can let π΄ be the point π₯ one,
π¦ one and π΅ be the point π₯ two, π¦ two and then substitute the values for these
coordinates into the slope formula. That gives five minus two in the
numerator and zero minus negative 10 in the denominator. And we need to be especially
careful here to ensure we include both of those negative signs.

That simplifies to three over 10
because zero minus negative 10 in the denominator is zero plus 10, which is 10. Substituting π equals three-tenths
into the slopeβintercept form then, we have π¦ equals three-tenths of π₯ plus
five. We need to rearrange this equation,
though, so that it is in the general form. And to do this, we need to collect
all of the terms on the same side of the equation. Subtracting three-tenths of π₯ and
five from each side, we have negative three-tenths of π₯ plus π¦ minus five equals
zero.

Now itβs also usual that in the
general form, the values of π, π, and π are integers, so weβll multiply the
entire equation by 10, which gives negative three π₯ plus 10π¦ minus 50 equals
zero. And this is in the general form
with π equal to negative three, π equal to 10, and π equal to negative 50. Now we could have grouped the terms
on the other side of the equation if after this stage here we had subtracted π¦ from
each side and then multiplied by 10. This wouldβve given three π₯ minus
10π¦ plus 50 equals zero, which is the exact negative of our equation, and therefore
an alternative answer also in its general form. Our answer is negative three π₯
plus 10π¦ minus 50 equals zero.

In our final example, weβll find
the equation of a straight line in the general form, this time given its π₯- and
π¦-intercept.

Determine the equation of the line
which cuts the π₯-axis at four and the π¦-axis at seven.

Although it isnβt entirely
necessary, we may find it helpful to sketch this straight line. It cuts the π₯-axis at four, and it
cuts the π¦-axis at seven. So it looks something like
this. We can see that this line slopes
downwards from left to right, and so it has a negative slope. This will give us one way of
checking whether our answer, when we reach it, seems reasonable. To answer this question, weβll
begin by using the slopeβintercept form of the equation of a straight line: π¦
equals ππ₯ plus π, where π is the slope of the line and π is its
π¦-intercept. We already know the value of π
because it was given in the question. We were told the line cuts the
π¦-axis at seven. So our line is π¦ equals ππ₯ plus
seven. And we need to calculate its
slope.

The slope of a straight line
passing through the two points with coordinates π₯ one, π¦ one and π₯ two, π¦ two is
π¦ two minus π¦ one over π₯ two minus π₯ one. Using the coordinates of the π₯-
and π¦-intercepts then, which are four, zero and zero, seven, we can calculate the
slope of this line. Itβs zero minus seven over four
minus zero, which simplifies to negative seven over four. This is a negative value, so we can
be reassured that what weβve done so far is correct. The equation of this line, then, is
π¦ equals negative seven over four π₯ plus seven.

Now we could leave our answer in
this form, but it looks a little unfriendly because we have a quotient. We can find an equivalent form of
this equation by rearranging to the general form. Weβll begin by multiplying every
term in the equation by four, giving four π¦ equals negative seven π₯ plus 28. Next, weβll group all of the terms
on the same side of the equation, choosing the left-hand side so that the
coefficients of both π₯ and π¦ are positive. So adding seven π₯ and subtracting
28 from each side, we have seven π₯ plus four π¦ minus 28 is equal to zero. This is the equation of this
straight line in its most general form. Thatβs ππ₯ plus ππ¦ plus π
equals zero, where π, π, and π are real constants.

Letβs now summarize the key points
from this video. The general form of the equation of
a straight line is ππ₯ plus ππ¦ plus π equals zero, where π, π, and π are real
constants, which are usually integers. This is called the general form
because every straight line, whether itβs diagonal, vertical, or horizontal, can be
represented by an equation in this form. Provided π is nonzero, the slope
of a line given in general form is negative π over π, and the π¦-intercept is
negative π over π. We also saw that we can find the
equation of a straight line in the general form given different sets of information
about the line and that we can rearrange the equation of a straight line given in
any other form to the general form using algebraic manipulation.