### Video Transcript

In this video, we will learn how to
find the intersection point of two straight lines on the coordinate plane and use
this concept to find equations of lines. Letβs begin by thinking about what
we mean by the intersection point. We say that the intersection of two
lines is the single point where they meet or cross each other. So we could, for example, have this
line segment π΄π΅ and another line segment ππ. We could then mark on the point πΉ
where these two line segments cross. Therefore, point πΉ is the
intersection of line segment π΄π΅ and line segment ππ. So when it comes to finding the
intersection of two lines on a coordinate plane, the same principle still
applies. Weβre looking to find the point
where the two lines meet or cross.

Weβll usually be given the
equations of two lines. So, the intersection point can be
given as an ordered pair or coordinate. We can find this ordered pair
graphically by drawing or algebraically by solving to find the π₯- and
π¦-values. Notice that because we usually have
two equations with two unknowns of π₯ and π¦, then weβll often need to solve
simultaneously or by using a substitution method. Weβll look at a number of different
methods of solving algebraically. But letβs have a look at the first
question where we use a graphical method.

At which point do the lines π₯
equals seven and one-sixth π¦ equals negative one intersect?

In this question, we have the
equations of two lines and weβre asked where these two lines intersect, which would
be the point where they meet or cross. It might be useful to begin
visualizing what these two lines would look like on the coordinate grid. The line π₯ equals seven indicates
all the ordered pairs that have an π₯-value of seven. Weβll have a vertical line that
goes through seven on the π₯-axis. For the second equation of a sixth
π¦ equals negative one, sometimes itβs more easy to visualize if we have it with π¦
as the subject. Rearranging by multiplying through
by six would give us the equation of π¦ equals negative six.

So the equation of the line π¦
equals negative six or sixth π¦ equals negative one will be a horizontal line
passing through negative six on the π¦-axis. The intersection point then is the
ordered pair where these two lines intersect. We can therefore give the answer of
the coordinate seven, negative six. Usually when we find the
intersection of two points, we could set the equations equal to each other. But as we had a horizontal and a
vertical line here, the only points where these two have the same π₯- and π¦-values
is when π₯ is equal to seven and π¦ is equal to negative six, giving us the
coordinate seven, negative six.

In the next few questions, weβll
use an algebraic method to find the intersection of two lines.

Determine the point of intersection
of the two straight lines represented by the equations π₯ plus three π¦ minus two
equals zero and negative π¦ plus one equals zero.

Letβs say to answer this question
weβre not going to draw these lines to get a graphical solution. Instead, weβre going to solve these
algebraically. At the point of intersection,
thatβs the place where the two lines meet or cross, the π₯- and π¦-values will be
the same. As we have two equations with the
two unknowns of π₯ or π¦, then weβre going to need to solve this simultaneously or
by using a substitution method. However, in our second equation, we
donβt actually have an π₯-value. So perhaps, a substitution method
here is the easiest. If we take our second equation of
negative π¦ plus one equals zero and rearrange this to make π¦ the subject, then by
adding π¦ to both sides, we would get one equals π¦ or π¦ equals one.

Now that weβve established that π¦
is equal to one, we can plug this into the first equation. This gives us π₯ plus three times
one subtract two equals zero. Evaluating this, we have π₯ plus
one equals zero. Subtracting negative one, we have
π₯ is equal to negative one. Now we know that at the point of
intersection of these two equations, the π₯-value is negative one and the π¦-value
is one, which means that we can give our answer as the coordinate negative one,
one.

In the next question, weβll find
the equation of a line passing through a point and the intersection point of another
two lines.

What is the equation of the line
passing through π΄ negative one, three and the intersection of the lines three π₯
minus π¦ plus five equals zero and five π₯ plus two π¦ plus three equals zero?

In this question, weβre really
interested in finding two points. Weβre given this point π΄ of
negative one, three, and then weβll find another point which is the intersection of
these two lines. So, the first step to take here is
to find the intersection of these lines. At the intersection point of two
lines, the π₯- and the π¦-values are the same. You might observe that we have two
unknowns to find, π₯ and π¦. But we do have two different
equations to help us solve it. One way to solve it is by solving
these two equations simultaneously. Itβs helpful to number our
equations equation one and equation two. And then we need to decide which
value of π₯ or π¦ to eliminate.

In order to eliminate either the
π₯- or π¦-value, their absolute values need to be the same. So if we choose to eliminate the
π¦-value, then weβll need to multiply all of equation one by two, which gives us six
π₯ minus two π¦ plus 10 equals zero. We can then put the second equation
underneath. In order to eliminate the two π¦βs
as we have a negative two π¦ and a positive two π¦, weβll be adding our two
equations. Six π₯ plus five π₯ gives us
11π₯. Negative two π¦ plus two π¦ does
indeed give us zero, which is what we wanted. 10 plus three gives us 13. And on the right-hand side, we have
zero. Now that we have 11π₯ plus 13
equals zero, we can rearrange this to find the value of π₯. Subtracting 13 from both sides and
then dividing through by 11 gives us that π₯ is equal to negative 13 over 11.

Now that we found the value of π₯,
we can plug this into either of our equations to find the value of π¦. Choosing the first equation would
give us three multiplied by negative 13 over 11 minus π¦ plus five equals zero. Simplifying the first value gives
us negative 39 over 11 and then we can add five. Remembering that five is equal to
fifty-five elevenths, weβll get the equation sixteen elevenths minus π¦ equals
zero. Adding π¦ to both sides gives us
sixteen elevenths equals π¦ or π¦ equals sixteen elevenths. Now that we found these π₯- and
π¦-values, it means that we know that the intersection point of our two lines is at
the point negative thirteen elevenths, sixteen elevenths.

Next, we need to find the equation
of the line that joins this point with the point π΄ negative one, three. Letβs start by finding the slope of
this line. The slope or π-value for any two
points π₯ one, π¦ one and π₯ two, π¦ two is given by π¦ two minus π¦ one over π₯ two
minus π₯ one. It doesnβt matter which coordinate
we designate with π₯ one, π¦ one values. So letβs put those values with the
point π΄. This gives us that the slope or π
is equal to sixteen elevenths subtract three over negative thirteen elevenths
subtract negative one. Simplifying the numerator, sixteen
elevenths subtract thirty-nine elevenths gives us negative seventeen elevenths. On the denominator, we have
negative thirteen elevenths plus one, and one is equivalent to eleven elevenths. So we have the answer on the
denominator of negative two elevenths.

Before we simplify this anymore, we
can eliminate the negative from both the numerator and denominator. We can then recognize that this
seventeen elevenths over two elevenths is equivalent to seventeen elevenths divided
by two elevenths. When weβre dividing fractions, itβs
equivalent to multiplying by the reciprocal of the second fraction. Taking out the common factor of 11
from the numerator and denominator before multiplying, we then get the answer that
π is equal to 17 over two. This means that the slope of the
equation joining our two points is 17 over two. Letβs clear some space for the next
part of our working.

Now that we have our two
coordinates and the slope of the line, we can use the point-slope form to find the
equation of the line. When we have the slope π of a line
and a coordinate π₯ one, π¦ one, then the equation is given by π¦ minus π¦ one
equals π times π₯ minus π₯ one. Given we have two points, we could
use either of these in the equation, but letβs use the coordinate of π΄ at negative
one, three with the π₯ one, π¦ one values. Plugging these into the formula, we
have π¦ minus three, as that was our π¦ one value, equals 17 over two, as that was
the π-value, multiplied by π₯ minus negative one, as that was the π₯ one value. On the right-hand side within the
parentheses, we have minus negative one, which is equivalent to plus one. So when we distribute the
parentheses on the right-hand side, we have 17 over two π₯ plus 17 over two.

As we want to collect together our
terms, then to collect the constant terms, weβll want to add three to both
sides. Three is equivalent to six over
two, so when we add that to 17 over two, we get 23 over two. At this point, we have a perfectly
valid answer for the equation of a line, but sometimes itβs nice to remove any
fractions if we can. So letβs multiply this equation
through by two. This gives us two π¦ equals 17π₯
plus 23. And then as an equation equal to
zero, weβd have zero equals 17π₯ minus two π¦ plus 23, which means that we can write
our answer in the slightly nicer format of 17π₯ minus two π¦ plus 23 equals
zero. And thatβs the equation of the line
passing through negative one, three and the intersection of the two lines three π₯
minus π¦ plus five equals zero and five π₯ plus two π¦ plus three equals zero.

Letβs have a look at another
question.

Find the equation of the straight
line that passes through the point of intersection of the two straight lines π₯
minus eight π¦ equals two and negative six π₯ minus eight π¦ equals one and is
parallel to the π¦-axis.

In this question, we need to find
the equation of a line which passes through the intersection point of two other
lines, and thatβs the point where these two lines would cross or meet. That wouldnβt be enough information
to find an equation, so weβre also told that this line is parallel to the
π¦-axis. The first step to take then is to
actually find the point of intersection between our two lines. At the point of intersection
between two lines, their π₯-values will be equal and their π¦-values will be
equal. We have two unknowns to find, the
π₯- and the π¦-values, but we also have two equations to help us. We can solve this algebraically
using substitution or by using simultaneous equations.

Using the simultaneous equations
method, weβd need to eliminate either the π₯- or π¦-values. But we might notice here that we
have negative eight π¦ in both of our equations. Therefore, if we subtracted our
second equation from our first equation, we could eliminate the π¦-value. Doing so would give us π₯ subtract
six π₯, which is seven π₯. Negative eight π¦ subtract negative
eight π¦ would give us zero, which is what we wanted. And finally, two subtract one gives
us one. Now that we know that seven π₯ is
equal to one, we can divide through by seven, which would give us π₯ is equal to
one-seventh. Now that we found the π₯-value, we
can plug this into either of our equations to find the value of π¦.

So, when π₯ is equal to
one-seventh, using the first equation, weβd have one-seventh subtract eight π¦ is
equal to two. Subtracting one-seventh from both
sides gives us negative eight π¦ equals two minus one-seventh. So on the right-hand side, weβll
have thirteen sevenths. Dividing both sides by negative
eight, we have π¦ is equal to 13 over negative 56, or π¦ is equal to negative 13
over 56. We have now found the two values of
π₯ and π¦, which means that the intersection point of the two straight lines we were
given is one-seventh, negative thirteen fifty-sixths.

We now need to find the equation of
the line which passes through this point and which is parallel to the π¦-axis. It might be useful to think about
what this might look like on a coordinate grid. The point one-seventh, negative
thirteen fifty-sixths would be approximately here. The line going through this point
parallel to the π¦-axis will be a vertical line. But what exactly is the equation of
this line? All the values on this line will
have an π₯-value of one-seventh. Therefore, the equation of the line
is π₯ equals one-seventh. And so weβve answered this question
by finding the intersection point of two lines and then working out of the equation
of this line given it was parallel to the π¦-axis.

We can now summarize some of the
key points of this video. Firstly, we saw that the
intersection point of two lines is the single point where they meet or cross each
other. We can find the intersection point
of two lines on the coordinate plane graphically by drawing the lines on a
coordinate grid or algebraically by solving simultaneously to find the π₯- and
π¦-values of the intersection point. An algebraic method will always
give us an exact solution, but it can be useful to incorporate graphical methods as
a check on our answer.