Question Video: Writing Linear Equations in Standard Form Mathematics • 8th Grade

Find the equation of the straight line that passes through points (1, 1) and (3, 4). Give your answer in standard form.

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Video Transcript

Find the equation of the straight line that passes through points one, one and three, four. Give your answer in standard form.

Equations for straight lines are usually written in the format π¦ equals ππ₯ plus π, where π is the slope of that line and π equals the π¦-intercept. That means weβll need both the slope and the π¦-intercept to write this equation. The formula for the slope between two points is π¦ two minus π¦ one over π₯ two minus π₯ one. The changes in π¦ over the changes in π₯. Letβs start there.

We can label our two points: our first point π₯ one, π¦ one and our second point π₯ two, π¦ two. We plug in the information weβre given. π¦ two minus π¦ one becomes four minus one. And π₯ two minus π₯ one becomes three minus one. Four minus one is three. Three minus one is two. Our slope is three over two. Letβs use slope intercept form to find what the π¦-intercept is.

π¦ equals three over two π₯ plus π. Weβre going to take one of our two points and plug them in for π₯ and π¦. It doesnβt matter which of the two points we use as long as weβre consistent. We have to use the π₯- and π¦-coordinate from the same point. If we use π₯ two, π¦ two, four is equal to three-halves times three plus π. Three-halves times three equals nine-halves. Bring down the plus π. Bring down the four.

Weβre trying to solve for π which means we need to move the nine-halves to the other side of the equation. We subtract nine-halves on both sides. Four minus nine-halves could be written as eight-halves minus nine-halves. Eight-halves minus nine-halves is negative one-half, equals π.

Back to the slope intercept form, we plug in three-halves. And the π¦-intercept, negative one-half, is plugged in for π. We have to be careful here because π¦-intercept form is not standard form of a linear equation. A linear equation written in standard form follows the pattern ππ₯ plus ππ¦ equals π. So now itβs our job to take the slope intercept form and turn it into standard form to get the constant value by itself.

I can do that by subtracting three-halves π₯ from both sides of the equation. On the right, they cancel out leaving us with negative one-half equals π¦ minus three-halves π₯. Weβll just bring it up, so we have a little bit more room. Something else we need to know about π, π, and π is that these values need to be integers. They shouldnβt be fractions. Both our π- and π-values are given as fractions. But if we multiply both sides of our equation by two, we can get rid of these fractions.

π¦ times two equals two π¦. Three-halves times two equals three. Negative one-half times two equals negative one. The next thing we notice about our standard form is that weβre adding ππ₯ plus ππ¦. In our equation, we are subtracting π¦ from π₯. We want π₯ to be leading. And we want it to be positive. We want that π-value to be positive. So we multiply both sides of our equation by negative one. Negative one times two π¦ equals negative two π¦. Negative one times negative three π₯ equals positive three π₯. Negative one times negative one equals one. Our last step here will be to switch our π₯ and π¦.

Now, we have three π₯ minus two π¦ equals one. Standard form of a straight line that passes through one, one and three, four is three π₯ minus two π¦ equals one. Remember, what weβre looking for here is a positive π₯. And π, π, and π must be integers. They cannot be fractions.