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In what ratio does the point 24 over 11 π¦ divide the line segment joining the points π: two, negative two and π: three, seven?
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Also, find the value of π¦.
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In these sorts of questions, itβs always good to just sketch things out to get a rough idea of whatβs going on.
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In this case, weβve got point π down here at two, negative two and π up here at three, seven.
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And the other point letβs call that point π
.
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The other point will be between π and π because its π₯-coordinate, 24 over 11, is bigger than two but less than three.
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Now from this sketch, we donβt actually know whether this is going to be below the π₯-axis or above the π₯-axis.
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But weβll worry about that later.
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So in the question, weβre looking for the ratio ππ
to π
π.
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Now ππ
π is a straight line, so the π¦-coordinates change at a constant rate or the line has a constant slope or gradient.
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Now we were given the π₯-coordinates for all three points.
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So if we find the ratio of the differences in the π₯-coordinates from π to π
to π
to π, thatβll be the same as the ratios of the differences in the π¦-coordinates and, what weβre looking for, the ratio of the lengths of the line segments ππ
to π
π.
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So looking at the π₯-coordinates of π and π
, weβre going from an π₯-coordinate of two up to an π₯-coordinate of 24 over 11.
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So to work out the difference in those π₯-coordinates, we simply subtract the first from the second.
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And subtraction of fractions is much easier when theyβve got a common denominator.
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So instead of writing two, Iβm gonna write 22 over 11, because 22 over 11 is the same as two.
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Then 24 over 11 minus 22 over 11 is just two over 11.
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And the π₯-coordinates of points π
and π are 24 over 11 and three, so the difference in those is three minus 24 over 11.
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And again, to get the common denominator, we can reexpress three as 33 over 11.
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Then weβve got 33 over 11 minus 24 over 11, which is nine over 11.
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So the ratio ππ
to π
π is two elevenths to nine elevenths.
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But we wouldnβt normally leave those in fraction form.
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We can multiply both sides of that ratio by 11.
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And the 11s will cancel, enabling us to express that ratio, ππ
to π
π, in its simplest form as the lowest integer values that you can have for that ratio; thatβs two to nine.
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Now we need to find the value of π¦.
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And to do that, weβre gonna find the equation of the line that runs through points π and π.
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Now the general form of an equation of the straight line is π¦ is equal to ππ₯ plus π, where π is the gradient or slope of that line and π is the π¦-intercept, where it cuts the π¦-axis.
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Now we can work out the slope or the gradient by taking the difference in the π¦-coordinates and dividing that by the difference in the π₯-coordinates.
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So if we call the first coordinate pair π₯ one, π¦ one and the second coordinate pair π₯ two, π¦ two, π is defined as being π¦ two minus π¦ one over π₯ two minus π₯ one.
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So if we just substitute those values in for π₯ one, π₯ two, π¦ one, and π¦ two, we get seven minus negative two over three minus two.
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And seven take away negative two is nine, and three take away two is one.
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So the slope is nine over one, or just nine.
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So we now know that the equation of that line is gonna be π¦ is equal to nine π₯ plus wherever it cuts the π¦-axis.
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But we do know some points that are on the line.
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So we know that, for example, when π₯ is equal to three, then π¦ is equal to seven.
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So we can substitute those values in for π₯ and π¦ and work out the value of π.
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That means that seven is equal to nine times three plus π.
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Well, nine times three is 27.
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And if I now subtract 27 from both sides of that equation, Iβll find that negative 20 is equal to π, or π is equal to negative 20.
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And I can plug that in to my original π¦ equals ππ₯ plus π equation.
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So the equation of the line that goes through these two points is π¦ is equal to nine π₯ minus 20.
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All I now need to do is substitute in the π₯-coordinate, 24 over 11, and I can find out the corresponding π¦-coordinate.
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So π¦ is equal to nine times the π₯-coordinate, 24 over 11, minus 20.
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Well, nine times 24 over 11 is the same as nine times 20 over 11 plus nine times four over 11.
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And nine times 20 is 180, and nine times four is 36.
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So that first term becomes 180 plus 36 over 11, which is 216 over 11.
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Now rather than just subtracting 20, letβs turn that into a fraction that has a denominator of 11 as well.
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So we need to multiply 20 by 11.
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And that means we can express 20 as 220 over 11.
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Now weβve got 216 over 11 minus 220 over 11.
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And that tells us that the value of π¦ is negative four over 11.