Video Transcript
Determine the equation of the line
that passes through the points negative two, negative one and zero, three. Determine the equation of the line
that passes through the points negative two, four and negative one, one. Hence, do the lines intersect? If yes, state the point of
intersection.
We begin by recalling the general
form for the equation of a straight line. It’s 𝑦 equals 𝑚𝑥 plus 𝑐, where
𝑚 is the gradient and 𝑐 is the 𝑦-intercept. The alternative form we might use
is 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one. Here, 𝑚 is also the gradient. But instead of 𝑐, we substitute a
coordinate given by 𝑥 one, 𝑦 one in. We’re going to use the first form
of this equation. And so we also need to recall the
formula for the gradient of a straight line. It’s change in 𝑦 divided by change
in 𝑥 or rise over run. And it’s sometimes also written as
𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one.
Let’s label the point negative two,
negative one 𝑥 one, 𝑦 one and zero, three 𝑥 two, 𝑦 two. We could alternatively have chosen
to do this the other way around as long as we made sure that 𝑥 one, 𝑦 one and 𝑥
two, 𝑦 two are individually part of the same set of coordinates. Then, the gradient of this line is
three minus negative one over zero minus negative two. Three minus negative one is
four. And zero minus negative two is
two. So the gradient is four divided by
two, which is simply two.
So far, we can see that the
equation of our line is 𝑦 equals two 𝑥 plus 𝑐. But what is the value of 𝑐? What’s the 𝑦-intercept? Usually, we’d look to substitute a
pair of our coordinates into this equation and solve for 𝑐. However, if we look carefully, we
see that the line actually passes through the point zero, three. This actually lies on the 𝑦-axis
at three. And this means the 𝑦-intercept of
our line must be three. And therefore, its equation is 𝑦
equals two 𝑥 plus three.
Let’s repeat this process for our
second line. This time though, we’ll use the
alternative form of the equation of a straight line, 𝑦 minus 𝑦 one equals 𝑚 times
𝑥 minus 𝑥 one. Once again, we identify 𝑥 one, 𝑦
one and 𝑥 two, 𝑦 two. This time, the gradient is one
minus four over negative one minus negative two. One minus four is negative
three. And negative one minus negative two
is negative one plus two, which is one. So the gradient of this line is
negative three.
We’ll take our values of 𝑥 one, 𝑦
one to be negative two, four. But we could alternatively have
chosen negative one, one for the equation of a straight line. Our equation is therefore 𝑦 minus
four equals negative three times 𝑥 minus negative two. We’ll simplify this expression
before distributing our parentheses. And we see that 𝑦 minus four is
equal to negative three 𝑥 minus six. Our final step is to add four to
both sides of our equation. And we see the equation of our
straight line is 𝑦 equals negative three 𝑥 minus two.
The final part of this question
asks us whether the lines intersect. Well, let’s think about what it
would mean if the lines didn’t intersect, if they never met. We recall that the definition of
two lines are parallel is that they’ll never meet. And of course, two lines will be
parallel if their gradients are the same. We’ve already seen that the
gradient of our first line is two. And the gradient of our second line
is negative three. So they can’t be parallel. This means, yes, they will indeed
meet. So how do we find the points of
intersection? Well, to find the points of
intersection, we need to find a value of 𝑥 and 𝑦 that they have in common. That will be the point that they
meet. So we’re going to solve the
equation 𝑦 equals two 𝑥 plus three and 𝑦 equals negative three 𝑥 minus two
simultaneously. Let’s do that up here.
We know that, in our first
equation, 𝑦 is equal to two 𝑥 [plus] three. In our second equation, 𝑦 is equal
to negative three 𝑥 minus two. Since both expressions in 𝑥 are
equal to 𝑦, we can say that two 𝑥 plus three itself must be equal to negative
three 𝑥 minus two. We’ll begin to solve this equation
in 𝑥 by adding three 𝑥 to both sides. And we see that five 𝑥 plus three
is negative two. Next, we’ll subtract two. And we find that five 𝑥 is equal
to negative five. Finally, we divide through by
five. And we find 𝑥 to be equal to
negative one. So the lines intersect at the point
where the 𝑥-coordinate is equal to negative one. We substitute this value of 𝑥 into
either of our original equations to find the value for 𝑦. I’ve chosen the first one. So 𝑦 is equal to two times
negative one plus three, which is one. We can say that, yes, they do
indeed meet. They meet at the point negative
one, one.
