# Question Video: Finding the Equation of a Line given Two Points on the Line Mathematics • 8th Grade

A line πΏ passes through the points (1, 1) and (β5, β1). Work out the equation of the line, giving your answer in the form π¦ = ππ₯ + π.

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### Video Transcript

A line πΏ passes through the point one, one and negative five, negative one. Work out the equation of the line giving your answer in the form π¦ equals ππ₯ plus π. π is your slope and π is the π¦-intercept. To find the slope, thereβs a formula also known as π where we take π¦ two minus π¦ one over π₯ two minus π₯ one.

Now in order to find the π¦-intercept, if we had a graph, it would just be where we crossed the π¦-axis. Now since we donβt, once we find the slope, we can use the point one, one or negative five, negative one and plug it in for π₯ and π¦ and then we can solve for π at the end.

So letβs first begin by finding the slope of this line. So here we have our two points: one, one and negative five, negative one. And now weβll plug it into our formula. So we have π¦ two minus π¦ one, so negative one minus one, over π₯ two minus π₯ one, negative five minus one, which is negative two over negative six. And the two negatives cancel to make a positive and two-sixths reduces to one-third. So our slope is one-third.

So taking the equation of the line and plugging in one-third for π, now we have a π¦ and π₯ and a π left. Now the π¦ and π₯, we can actually substitute in one, one or negative five, negative one. Letβs go ahead and use the point one, one. So using one to plug in for π₯, one to plug in for π¦, and one-third to plug in for π β plugging one in for π₯ and one in for π¦ and one-third in for π β that means we have everything except for π. This allows us to solve for π. One-third times one is just one-third. So in order to solve for π, we need to subtract one-third from both sides of the equation. This means on the right-hand side, the one-third would cancel, but on the left we have one minus one-third.

In order to subtract fractions, they need to have a common denominator. So one is the same thing as three over three. So we can replace one with three-thirds. So three-thirds minus one-third, we subtract the numerators and we keep the denominators. So π is equal to two-thirds. Therefore, plugging in one-third for π, our slope, and two-thirds for π, our π¦-intercept, we get the equation π¦ equals one-third π₯ plus two-thirds.