Question Video: Finding the Equation of a Line given Two Points on the Line | Nagwa Question Video: Finding the Equation of a Line given Two Points on the Line | Nagwa

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

A line 𝐿 passes through the points (1, 1) and (βˆ’5, βˆ’1). Work out the equation of the line, giving your answer in the form 𝑦 = π‘šπ‘₯ + 𝑏.

02:34

Video Transcript

A line 𝐿 passes through the point one, one and negative five, negative one. Work out the equation of the line giving your answer in the form 𝑦 equals π‘šπ‘₯ plus 𝑏. π‘š is your slope and 𝑏 is the 𝑦-intercept. To find the slope, there’s a formula also known as π‘š where we take 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one.

Now in order to find the 𝑦-intercept, if we had a graph, it would just be where we crossed the 𝑦-axis. Now since we don’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 𝑦 and then we can solve for 𝑏 at the end.

So let’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 we’ll plug it into our formula. So we have 𝑦 two minus 𝑦 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‘š, now we have a 𝑦 and π‘₯ and a 𝑏 left. Now the 𝑦 and π‘₯, we can actually substitute in one, one or negative five, negative one. Let’s go ahead and use the point one, one. So using one to plug in for π‘₯, one to plug in for 𝑦, and one-third to plug in for π‘š β€” plugging one in for π‘₯ and one in for 𝑦 and one-third in for π‘š β€” that means we have everything except for 𝑏. This allows us to solve for 𝑏. One-third times one is just one-third. So in order to solve for 𝑏, 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 𝑏 is equal to two-thirds. Therefore, plugging in one-third for π‘š, our slope, and two-thirds for 𝑏, our 𝑦-intercept, we get the equation 𝑦 equals one-third π‘₯ plus two-thirds.

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