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 π¦ = ππ₯ + π. Given that the point (π₯β, π¦β) lies on the line, find an expression for π in terms of π, π₯β, and π¦β. Given also that the point (π₯β, π¦β) lies on the line and π is the slope of the line, find an expression for π in terms of π¦β, π¦β, π₯β, and π₯β. By substituting in for π and factorizing out π, find the formula for the equation of the line. Substitute in your expression for π to complete your formula.

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### Video Transcript

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

So, letβs start at the beginning. We can model a linear equation with the line π¦ equals ππ₯ plus π. If we know that the point π₯ one, π¦ one lies on this line, we can find an expression for π in terms of π, π₯ one, and π¦ one. If we want to find an expression for π in these terms, π will be on one side of the equation and the other side will only contain π, π₯ one, and π¦ one as those are the terms we want to know π in. If π₯ one, π¦ one lies on this line, then we can say that π¦ one is equal to π times π₯ one plus π.

And now, what we want is to rearrange the equation so that π is by itself. And that means weβll subtract ππ₯ one from both sides of the equation, which gives us π¦ one minus π times π₯ one equals π. But more commonly, weβll put the π on the left so that we have π equals π¦ one minus π times π₯ one. And we have our first expression, π¦ one minus π times π₯ one.

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

Remember, our goal is to get π by itself. So, we add π times π₯ two to both sides of the equation. And then, weβll subtract π¦ one from both sides of the equation to move that π¦ one value to the other side. On the left, weβll have π times π₯ two minus π times π₯ one since the π¦βs cancel out. And on the other side, weβll have π¦ two minus π¦ one since the ππ₯ two cancels out.

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

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

By substituting in for π and factoring out π, find the formula for the equation of the line.

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

Itβs also good to keep the π¦βs on the same side of the equation. And we can do that here by subtracting π¦ one from both sides, which gives us π¦ minus π¦ one equals π 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 into the equation from step three. Beginning here, π¦ minus π¦ one equals π times π₯ minus π₯ one. We take our value for π, π¦ two minus π¦ one over π₯ two minus π₯ one, and plug that in. Which gives us π¦ minus π¦ one equals π¦ two minus π¦ 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, π¦ one; π₯ two, π¦ two and solve for the equation.

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