Question Video: Deriving the Formula to Find the Equation of a Line | Nagwa Question Video: Deriving the Formula to Find the Equation of a Line | Nagwa

Question Video: Deriving the Formula to Find the Equation of a Line Mathematics

A straight line is defined by the equation 𝑦 = π‘šπ‘₯ + 𝑐. Given that the point (π‘₯₁, 𝑦₁) lies on the line, find an expression for 𝑐 in terms of π‘š, π‘₯₁, and 𝑦₁. Given also that the point (π‘₯β‚‚, 𝑦₂) lies on the line and π‘š is the slope of the line, find an expression for π‘š in terms of 𝑦₁, 𝑦₂, π‘₯₁, and π‘₯β‚‚. By substituting in for 𝑐 and factorizing out π‘š, find the formula for the equation of the line. Substitute in your expression for π‘š to complete your formula.

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Video Transcript

A straight line is defined by the equation 𝑦 equals π‘šπ‘₯ plus 𝑐. Given that the point π‘₯ one, 𝑦 one lies on the line, find an expression for 𝑐 in terms of π‘š, π‘₯ one, and 𝑦 one. Given also that the point π‘₯ two, 𝑦 two lies on the line and π‘š is the slope of the line, find an expression for π‘š in terms of 𝑦 one, 𝑦 two, π‘₯ one, and π‘₯ two. By substituting in for 𝑐 and factoring out π‘š, find the formula for the equation of the line. And finally, substitute in your expression for π‘š to complete your formula.

So, let’s start at the beginning. We can model a linear equation with the line 𝑦 equals π‘šπ‘₯ plus 𝑐. If we know that the point π‘₯ one, 𝑦 one lies on this line, we can find an expression for 𝑐 in terms of π‘š, π‘₯ one, and 𝑦 one. If we want to find an expression for 𝑐 in these terms, 𝑐 will be on one side of the equation and the other side will only contain π‘š, π‘₯ one, and 𝑦 one as those are the terms we want to know 𝑐 in. If π‘₯ one, 𝑦 one lies on this line, then we can say that 𝑦 one is equal to π‘š times π‘₯ one plus 𝑐.

And now, what we want is to rearrange the equation so that 𝑐 is by itself. And that means we’ll subtract π‘šπ‘₯ one from both sides of the equation, which gives us 𝑦 one minus π‘š times π‘₯ one equals 𝑐. But more commonly, we’ll put the 𝑐 on the left so that we have 𝑐 equals 𝑦 one minus π‘š times π‘₯ one. And we have our first expression, 𝑦 one minus π‘š times π‘₯ one.

We also know that the point π‘₯ two, 𝑦 two lies on this line and π‘š is the slope of the line. We wanna find an expression for π‘š in terms of 𝑦 one, 𝑦 two, π‘₯ one, and π‘₯ two. And similarly, if we want π‘š in terms of these values, π‘š needs to be by itself. And the other side of the equation will only include 𝑦 one, 𝑦 two, π‘₯ one, and π‘₯ two. From part one, we know that 𝑐 equals 𝑦 one minus π‘š times π‘₯ one. And since π‘₯ two, 𝑦 two also lies on this line, we can plug in π‘₯ two and 𝑦 two into this equation. Since these are at the same lines, their 𝑐-values will be the same. And that means we can take 𝑦 one minus π‘š times π‘₯ one and set it equal to 𝑦 two minus π‘š times π‘₯ two.

Remember, our goal is to get π‘š by itself. So, we add π‘š times π‘₯ two to both sides of the equation. And then, we’ll subtract 𝑦 one from both sides of the equation to move that 𝑦 one value to the other side. On the left, we’ll have π‘š times π‘₯ two minus π‘š times π‘₯ one since the 𝑦’s cancel out. And on the other side, we’ll have 𝑦 two minus 𝑦 one since the π‘šπ‘₯ two cancels out.

To get π‘š by itself, we see that we have two π‘š-values that we can factor out, which will give us π‘š times π‘₯ two minus π‘₯ one equals 𝑦 two minus 𝑦 one. From there to finally get π‘š by itself, we’ll divide both sides of the equation by π‘₯ two minus π‘₯ one. On the left, they cancel out, and we have π‘š equals 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one.

This completes part two. So, we clear some room for part three.

By substituting in for 𝑐 and factoring out π‘š, find the formula for the equation of the line.

This time, we start back with the definition of a straight line 𝑦 equals π‘šπ‘₯ plus 𝑐. And in place of 𝑐, we plug in 𝑦 one minus π‘š times π‘₯ one and everything else remains the same. Since we’re only adding and subtracting each of the terms, we can take away the parentheses. Our instructions say we need to factor out π‘š. To make that easier to see, we can put the order 𝑦 one plus π‘šπ‘₯ minus π‘š times π‘₯ one. And now, we’re able to factor out the π‘š so that we have 𝑦 equals 𝑦 one plus π‘š times π‘₯ minus π‘₯ one.

It’s also good to keep the 𝑦’s on the same side of the equation. And we can do that here by subtracting 𝑦 one from both sides, which gives us 𝑦 minus 𝑦 one equals π‘š times π‘₯ minus π‘₯ one. This is our equation for the line from step three.

In our final step, we want to substitute in what we found for π‘š into the equation from step three. Beginning here, 𝑦 minus 𝑦 one equals π‘š times π‘₯ minus π‘₯ one. We take our value for π‘š, 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one, and plug that in. Which gives us 𝑦 minus 𝑦 one equals 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one times π‘₯ minus π‘₯ one. And this completes our formula.

For any linear equation, if you know two points that fall on that line, you can plug those values in for π‘₯ one, 𝑦 one; π‘₯ two, 𝑦 two and solve for the equation.

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