### Video Transcript

In this video, we will learn how to
find the equation of a straight line in the form π¦ minus π¦ one equals π π₯ minus
π₯ one or ππ₯ plus ππ¦ plus π equals zero, given the slope and one point or two
different points that it passes through. An equation of a straight line in
two variables is any polynomial equation of degree one containing exactly two
variables, which are usually π₯ and π¦. We have already encountered
equations of straight lines in the slopeβintercept form, π¦ equals ππ₯ plus π. Here, π is the slope of the line
and π is the value of the π¦-intercept.

In this video, we will introduce
two new forms in which the equation of a straight line can be given and learn how to
derive these two equations given two pieces of data: either the coordinates of two
points that lie on a line or the coordinates of a single point and the lineβs
slope.

Letβs consider first the
pointβslope form of the equation of a straight line, π¦ minus π¦ one equals π π₯
minus π₯ one. Here, π is the slope of the line
and π₯ one, π¦ one are the coordinates of any given point that lies on the line. One way to think about this form of
the equation of a straight line is as a reformulation of the definition of the
slope. The slope of a straight line is
defined as the change in π¦ divided by the change in π₯. If we take π₯, π¦ to be the general
point on the line and π₯ one, π¦ one to be a specific point whose coordinates we
know, then we can write the slope π as π¦ minus π¦ one over π₯ minus π₯ one. Multiplying both sides of this
equation by π₯ minus π₯ one and then swapping the two sides of the equation around
gives the pointβslope form of the equation of a straight line: π¦ minus π¦ one
equals π π₯ minus π₯ one.

If we know the value of π and the
coordinates of one point on the line, we can then substitute these values into this
formula to find the equation of any given line. If instead we know the coordinates
of two points on the line, we can first calculate the value of π by applying the
slope formula and then substitute this and the coordinates of either point to find
the equation of the line.

Letβs now consider an example in
which we find the pointβslope form of the equation of a straight line, given its
slope and the coordinates of a point on the line.

Find in pointβslope form the
equation of the graph with slope four that passes through the point two, negative
three.

Weβre asked to find the equation of
this straight line in pointβslope form, which we recall is π¦ minus π¦ one equals π
π₯ minus π₯ one, where π is the slope of the line and π₯ one, π¦ one are the
coordinates of any point on the line. Weβre told that the slope of the
line is four β so this is the value of π β and that the line passes through the
point two, negative three, so these are the coordinates π₯ one, π¦ one. Substituting these values gives π¦
minus negative three is equal to four multiplied by π₯ minus two.

We can simplify the left-hand side
of the equation as π¦ minus negative three is equal to π¦ plus three. But as weβve been asked to give the
answer in pointβslope form, we wonβt rearrange the equation any further. Hence, weβve found that the
equation of the graph with slope four that passes through the point two, negative
three, in pointβslope form, is π¦ plus three equals four π₯ minus two.

Both the pointβslope and
slopeβintercept forms are useful forms of the equation of a straight line and
immediately reveal properties such as the slope or π¦-intercept of a straight
line. But there is a major drawback of
both of these forms, which is that not every equation of a straight line can be
written in these forms. Specifically, consider lines that
are parallel to the π¦-axis, which all have equations of the form π₯ equals some
constant π. It is not possible to rearrange
either of the forms weβve discussed so far to include this category of straight
lines. And so, we introduce a third form,
the general form of the equation of a straight line: ππ₯ plus ππ¦ plus π equals
zero, where π, π, and π are constants. Note that the value of π here is
not the same as the value of π in the slopeβintercept form.

The general form has the
significant advantage that the equation of every straight line, whether it is
vertical, horizontal, or diagonal, can be written in this form. It is also straightforward to
convert between the general and slopeβintercept forms. This can be achieved by subtracting
both ππ₯ and π from both sides of the equation of a line given in general form and
then dividing both sides by π, provided π is nonzero. The slope and π¦-intercept can then
be easily identified from this rearranged form.

Letβs now consider an example in
which we find the equation of a line in the general form given the coordinates of
two points that lie on a line.

A line πΏ passes through the points
three, three and negative one, zero. Work out the equation of the line,
giving your answer in the form ππ₯ plus ππ¦ plus π equals zero.

Weβve been given the coordinates of
two points that lie on line πΏ and asked to find its equation in general form. Weβll begin by calculating the
slope of the line, using the formula π equals π¦ two minus π¦ one over π₯ two minus
π₯ one, or change in π¦ over change in π₯. Substituting the coordinates of the
two points gives π equals zero minus three over negative one minus three, which
simplifies to negative three over negative four and then three-quarters.

We can then recall that if we know
the slope of a line and the coordinates of any point on it, we can find its equation
by first using the pointβslope form of the equation of a straight line: π¦ minus π¦
one equals π π₯ minus π₯ one. We can use either of the two points
and the choice is entirely arbitrary. Substituting π equals
three-quarters and both π₯ one and π¦ one equal three gives π¦ minus three equals
three-quarters of π₯ minus three.

Now we need to rearrange this
equation into general form. Multiplying both sides by four to
eliminate the fractions gives four π¦ minus 12 equals three multiplied by π₯ minus
three. Next, expanding the brackets on the
right-hand side gives four π¦ minus 12 equals three π₯ minus nine. We now need to collect all the
terms on the same side of the equation. So subtracting three π₯ and adding
nine to both sides gives negative three π₯ plus four π¦ minus three is equal to
zero. Equally, we could have collected
all the terms on the other side of the equation, which would have resulted in the
exact negative of this expression. But as it is equal to zero, the
equation is equivalent.

Itβs also worth pointing out that
if we had chosen to substitute the coordinates of the other given point, negative
one, zero, we would have arrived at the same answer once the equation was fully
rearranged. Weβve found the equation of the
line that passes through the points three, three and negative one, zero in general
form is negative three π₯ plus four π¦ minus three is equal to zero.

In the previous example, we
multiplied the entire equation through by four, so that the coefficients in our
final answer were all integers. This is generally what we do when
given the equation of a straight line in general form. Whilst not strictly essential, it
is usual for π, π, and π to be the smallest possible integers. We eliminate any fractions by
multiplying by the denominators. And then if the coefficients all
share a common factor, we can divide the entire equation through by this common
factor to give the equation in its simplest form. Our answer isnβt incorrect if we
donβt do this, but as there are infinitely many alternative equivalent equations, it
is usual to give the final answer in this form.

Weβll now consider some other
examples, in which weβll apply the skills weβve developed to more complex problems
involving the equation of a straight line. In the first of these, weβll find
the equation of a straight line given its slope and the fact that it shares its
π₯-intercept with another straight line.

The line π¦ equals four π₯ minus
eight crosses the π₯-axis at the point π΄. Find the equation of the line with
a slope of negative three passing through the point π΄.

Weβre given the equation of a
straight line in slopeβintercept form and told that it crosses the π₯-axis at point
π΄. We can deduce that this line has a
slope of four and a π¦-intercept of negative eight, although this isnβt entirely
necessary. To find the coordinates of point
π΄, we recall that the π¦-coordinate of any point on the π₯-axis is zero. So, substituting π¦ equals zero
into the equation of this straight line gives an equation we can solve to find the
π₯-coordinate of point π΄. Adding eight to both sides of the
equation and then dividing by four gives π₯ equals two.

At this point, our sketch gives
some reassurance that our working so far is correct as the π₯-intercept occurs at a
positive π₯-value. We now know that the line whose
equation we want to determine has a slope of negative three and passes through the
point π΄, which has coordinates two, zero. We can find this equation by
substituting these values into the pointβslope form of the equation of a straight
line: π¦ minus π¦ one equals π π₯ minus π₯ one. Doing so gives π¦ minus zero equals
negative three π₯ minus two. We can simplify the left-hand side
of the equation to simply π¦. And as weβre not asked to give the
answer in a particular form, we can stop here and leave our answer in pointβslope
form. The equation of the given straight
line is π¦ equals negative three multiplied by π₯ minus two.

Equally, we could expand the
brackets and give our answer in slopeβintercept form as π¦ equals negative three π₯
plus six. Or we could rearrange the equation
to the general form and give our answer as three π₯ plus π¦ minus six is equal to
zero. All of these forms are equivalent
and correct.

In our next example, weβll find the
equation of a line given the coordinates of one point that lies on it and the
knowledge that the line passes through the intersection point of two other
lines.

The lines π¦ equals negative
one-third π₯ plus six and π¦ equals π₯ minus two intersect at the point π. The point π has coordinates
negative five, eight. Find the equation of the line
passing through the points π and π.

Weβre given the coordinates of one
point on the line, point π, and told that the line also passes through point π,
which is the point of intersection of two other straight lines. To find the coordinates of point
π, we need to solve the equations of the two straight lines simultaneously. As the two equations are both of
the form π¦ equals some function of π₯, we can eliminate the π¦-variable by equating
the two expressions on the other side of the equation to give π₯ minus two equals
negative one-third π₯ plus six. Multiplying both sides of this
equation by three gives three π₯ minus six equals negative π₯ plus 18. And then collecting like terms by
adding π₯ and six to both sides gives four π₯ equals 24. Finally, we can divide both sides
of the equation by four to find that the π₯-coordinate of point π is six.

To find the π¦-coordinate, we can
substitute π₯ equals six into either equation. But it will be simpler to use the
equation that doesnβt involve fractions. Substituting π₯ equals six into the
equation π¦ equals π₯ minus two, we find that the π¦-coordinate of point π is
four.

So, we now know the coordinates of
both points that lie on the line whose equation we wish to find. We can find the slope of the line
using the formula π equals change in π¦ over change in π₯, or π¦ two minus π¦ one
over π₯ two minus π₯ one. Substituting the coordinates of π
and π gives π equals eight minus four over negative five minus six, which is four
over negative 11 or negative four elevenths. We can then find the equation of
the straight line using the pointβslope form: π¦ minus π¦ one equals π π₯ minus π₯
one. We can substitute the coordinates
of either point π or π for π₯ one, π¦ one. But it probably makes more sense to
use point π as we were given these coordinates explicitly. We obtain π¦ minus eight equals
negative four elevenths π₯ minus negative five.

Weβre not asked to give our answer
in a particular form, so we can leave it as it is. The only simplification weβll make
is to rewrite π₯ minus negative five as π₯ plus five. Weβve found that the equation of
the line passing through points π and π is π¦ minus eight equals negative four
elevenths π₯ plus five.

Letβs finish by summarizing the key
points from this video. Given the coordinates of any two
points that lie on a straight line, π₯ one, π¦ one and π₯ two, π¦ two, we can find
the slope of that line using the formula change in π¦ over change in π₯, or π¦ two
minus π¦ one over π₯ two minus π₯ one. We recalled the slopeβintercept
form of the equation of a straight line, π¦ equals ππ₯ plus π, where π represents
the slope and π represents the π¦-intercept of the line. We then introduced a new form of
the equation of a straight line, the pointβslope form, π¦ minus π¦ one equals π π₯
minus π₯ one.

Given the slope π of a line and
the coordinates π₯ one, π¦ one of any point that lies on that line, we can find the
equation of the line by substituting π, π₯ one, and π¦ one into the pointβslope
form of the equation of a straight line. We then saw that whilst the first
two forms of the equation of a straight line are useful in certain contexts, they
both have the major drawback that they do not include all straight lines. And in particular, the equations of
vertical lines cannot be expressed in either of these forms. We therefore introduced the general
form of the equation of a straight line: ππ₯ plus ππ¦ plus π equals zero, which
has the significant advantage that the equation of any straight line can be written
in this form.

Finally, for those lines whose
equations can be written in multiple forms, we can convert between these different
forms by rearranging the equation.