WEBVTT
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Find the equation of the straight line that passes through points one, one and three, four.
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Give your answer in standard form.
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Equations for straight lines are usually written in the format π¦ equals ππ₯ plus π, where π is the slope of that line and π equals the π¦-intercept.
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That means weβll need both the slope and the π¦-intercept to write this equation.
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The formula for the slope between two points is π¦ two minus π¦ one over π₯ two minus π₯ one.
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The changes in π¦ over the changes in π₯.
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Letβs start there.
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We can label our two points: our first point π₯ one, π¦ one and our second point π₯ two, π¦ two.
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We plug in the information weβre given.
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π¦ two minus π¦ one becomes four minus one.
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And π₯ two minus π₯ one becomes three minus one.
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Four minus one is three.
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Three minus one is two.
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Our slope is three over two.
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Letβs use slope intercept form to find what the π¦-intercept is.
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π¦ equals three over two π₯ plus π.
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Weβre going to take one of our two points and plug them in for π₯ and π¦.
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It doesnβt matter which of the two points we use as long as weβre consistent.
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We have to use the π₯- and π¦-coordinate from the same point.
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If we use π₯ two, π¦ two, four is equal to three-halves times three plus π.
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Three-halves times three equals nine-halves.
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Bring down the plus π.
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Bring down the four.
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Weβre trying to solve for π which means we need to move the nine-halves to the other side of the equation.
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We subtract nine-halves on both sides.
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Four minus nine-halves could be written as eight-halves minus nine-halves.
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Eight-halves minus nine-halves is negative one-half, equals π.
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Back to the slope intercept form, we plug in three-halves.
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And the π¦-intercept, negative one-half, is plugged in for π.
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We have to be careful here because π¦-intercept form is not standard form of a linear equation.
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A linear equation written in standard form follows the pattern ππ₯ plus ππ¦ equals π.
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So now itβs our job to take the slope intercept form and turn it into standard form to get the constant value by itself.
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I can do that by subtracting three-halves π₯ from both sides of the equation.
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On the right, they cancel out leaving us with negative one-half equals π¦ minus three-halves π₯.
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Weβll just bring it up, so we have a little bit more room.
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Something else we need to know about π, π, and π is that these values need to be integers.
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They shouldnβt be fractions.
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Both our π- and π-values are given as fractions.
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But if we multiply both sides of our equation by two, we can get rid of these fractions.
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π¦ times two equals two π¦.
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Three-halves times two equals three.
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Negative one-half times two equals negative one.
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The next thing we notice about our standard form is that weβre adding ππ₯ plus ππ¦.
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In our equation, we are subtracting π¦ from π₯.
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We want π₯ to be leading.
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And we want it to be positive.
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We want that π-value to be positive.
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So we multiply both sides of our equation by negative one.
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Negative one times two π¦ equals negative two π¦.
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Negative one times negative three π₯ equals positive three π₯.
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Negative one times negative one equals one.
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Our last step here will be to switch our π₯ and π¦.
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Now, we have three π₯ minus two π¦ equals one.
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Standard form of a straight line that passes through one, one and three, four is three π₯ minus two π¦ equals one.
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Remember, what weβre looking for here is a positive π₯.
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And π, π, and π must be integers.
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They cannot be fractions.