# Video: Equations of Straight Lines through Points and Real-World Simultaneous Equations

Determine the equation of the line that passes through the points (−2, −1) and (0, 3). Determine the equation of the line that passes through the points (−2, 4) and (−1, 1). Hence, do the lines intersect? If yes, state the point of intersection.

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### 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.