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In this video, we will learn how to find the coordinates of a point that divides a line segment on the coordinate plane in a given ratio using the section formula.
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Letβs begin by reviewing some important terminology, firstly, line segments.
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A line segment is a part of a line thatβs bounded by two distinct endpoints.
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For example, here we have the line segment π΄π΅.
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We can even consider a line segment on the coordinate plane.
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If we have the coordinates of π΄ and π΅, we can even find the midpoint of this line segment by using the midpoint formula.
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In this video, however, we want to go one step further than simply dividing a line segment into two pieces.
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And weβll see how we can find the point which divides a line segment into a given ratio.
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Letβs see what that would look like.
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Here we have a line segment π΄π΅, and weβre given the coordinates of these two points.
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Letβs then say that we need to find the coordinates of this point π which divides line segment π΄π΅ in the ratio π to π.
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Is there a way in which we can find the coordinates of point π?
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Well, to do this, weβll need to start by constructing two right triangles.
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In these right triangles, one of them will have a hypotenuse of π and the other one will have a hypotenuse of π.
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Note that these two triangles will be similar since the corresponding angles of the triangle are congruent.
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In similar triangles, the ratio of corresponding sides will be equal.
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To find the distance of π from π΄, it will be π over π plus π multiplied by the length of π΄π΅.
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We can use this to find the π₯- and π¦-values of point π.
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So letβs begin with how we would find the π₯-value.
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π₯ is equal to π₯ sub one, thatβs the π₯-value in our π΄-point, plus π over π plus π multiplied by π₯ sub two subtract π₯ sub one.
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This second part of this equation represents our distance of π from point π΄ in terms of the π₯-values.
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In order to simplify this, we now need to make sure that our value of π₯ one is also as a fraction over π plus π.
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If we multiply π₯ sub one by π plus π on the numerator and denominator, weβll get π₯ is equal to π plus π times π₯ sub one over π plus π plus π multiplied by π₯ sub two subtract π₯ sub one over π plus π.
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We can then expand both sets of parentheses, and we notice that we have ππ₯ sub one subtract ππ₯ sub one.
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Weβre then left with π₯ is equal to ππ₯ sub two plus ππ₯ sub one over π plus π.
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We could then perform exactly the same process to find the value of π¦.
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This time, weβd swap our π₯ sub one and π₯ sub two values for π¦ sub one and π¦ sub two, respectively.
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We would then find that π¦ is equal to ππ¦ sub two plus ππ¦ sub one over π plus π.
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And so weβve now found the π₯-value and the π¦-value of point π, which divides our line segment.
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Letβs write this in a nicer way so that we can make a note of it.
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If π΄ with coordinates π₯ sub one, π¦ sub one and π΅ with coordinates π₯ sub two, π¦ sub two and point π divides line segment π΄π΅ such that π΄π to ππ΅ equals the ratio π to π, then π has coordinates.
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π equals ππ₯ sub two plus ππ₯ sub one over π plus π, ππ¦ sub two plus ππ¦ sub one over π plus π.
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We can now see how our π₯- and π¦-values come together to create this coordinate of π.
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This formula does look quite difficult, but when we put it into practice, itβs really not so difficult.
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Itβs not the easiest formula to remember, but as we go through this video, weβll write it out in each question.
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And hopefully, by the end of it, weβll have learned it a little bit more.
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Letβs take a look at our first question.
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If the coordinates of π΄ and π΅ are five, five and negative one, negative four, respectively, find the coordinates of the point πΆ that divides vector π΄π΅ internally by the ratio two to one.
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It might be sensible to begin a question like this by plotting our two coordinates.
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We could either create a sketch of this graph or use grid paper.
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So here we have a grid drawn.
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The point π΄ is at the coordinate five, five and π΅ is at negative one, negative four.
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As weβre told that thereβs a vector π΄π΅, we can join our two points.
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Weβre then asked to find the coordinates of a point πΆ which divides our vector π΄π΅ internally in the ratio two to one.
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This means that weβll need to use the section formula.
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For point π΄ with coordinates π₯ sub one, π¦ sub one and point π΅ with coordinates π₯ sub two, π¦ sub two, the point π which divides line segment π΄π΅ in the ratio π to π has coordinates π is equal to π times π₯ sub two plus π times π₯ sub one over π plus π, π times π¦ sub two plus π times π¦ sub one over π plus π.
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The key pieces of information that we need for the section formula are the two coordinates of π΄ and π΅ and the ratio.
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As the direction is important, weβre going from π΄ to π΅, then our π₯ sub one, π¦ sub one values must go with point π΄.
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The ordering of our ratio values of π and π is also important, so π will be two and π will be one.
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We can then plug these values into the formula and simplify.
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For the π₯-value of point π, we have two multiplied by negative one which is negative two plus five gives us three, and two plus one on the denominator of course becomes three.
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For the π¦-value, we have two multiplied by negative four, which is negative eight, plus one times five, which is five, gives us a value of negative three on the numerator.
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And our denominator will also be three.
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Simplifying these two fractions then, three over three is one and negative three over three is negative one.
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So now we know that point πΆ which divides this vector π΄π΅ in the ratio two to one is the coordinate one, negative one.
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A really good check of the answer if we have drawn a graph on grid paper is to check that the coordinate actually does lie on the line, and it does.
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Here we can see point πΆ with coordinates one, negative one.
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We can also see that line segment π΄π΅ is divided in the ratio two to one.
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Weβve therefore verified the coordinate one, negative one is the answer.
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Letβs take a look at another question.
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The coordinates of π΄ and π΅ and are one, nine and nine, nine, respectively.
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Determine the coordinates of the points that divide line segment π΄π΅ into four equal parts.
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Letβs start this question by imagining this line segment joining π΄ and π΅.
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We can consider this line split into four parts.
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And if we wanted to find this first point, we could consider how we would split this line segment π΄π΅ in a ratio one to three.
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The second point could be found by splitting it in the ratio two to two or even by finding the midpoint.
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We could find the coordinates of the third point by splitting the line segment π΄π΅ in the ratio three to one.
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We could actually do these three pieces of working using the section formula.
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This tells us that for any two points π΄ and π΅ with coordinates π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two, respectively, the point π which divides line segment π΄π΅ in the ratio π to π has coordinates π equal to ππ₯ sub two plus ππ₯ sub one over π plus π, ππ¦ sub two plus ππ¦ sub one over π plus π.
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In this question, we need to apply the section formula three times to find these three different coordinates with their different ratios.
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There is actually an easy way if we consider these two coordinates one, nine and nine, nine.
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We can plot these as shown and even create the line segment π΄π΅.
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As this is a horizontal line, itβs a little bit easier to work out the distance of π΄π΅.
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It would be eight units long.
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We can then divide this length into four equal parts, and so weβll get the coordinates three, nine; five, nine; and seven, nine.
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This will be our answer for the coordinates that divide line segment π΄π΅ into four equal parts.
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We could have done this in the same way using the section formula, but it would have taken a lot longer.
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Letβs take a look at another question.
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If πΆ is an element of π΄π΅ and vector π΄π΅ equals three times vector πΆπ΅, then πΆ divides vector π΅π΄ by the ratio blank.
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Option (A) two to one, option (B) one to two, option (C) one to three, option (D) three to one.
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Thereβs quite a lot of information in this question.
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But letβs start with the fact that we have this line segment π΄π΅, which we could model like this.
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Weβre told that πΆ is an element of π΄π΅, so that means that there will be a point πΆ somewhere on this line.
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If vector π΄π΅ is equal to three times vector πΆπ΅, then that means that three of this length πΆπ΅ would make up the length of π΄π΅.
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We could divide our length π΄π΅ into three pieces, but the question is, is πΆ here or is πΆ here?
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If we consider if πΆ is at this lower point, then the length πΆπ΅ would look like this.
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But if we were to multiply πΆπ΅ by three, we wouldnβt get the length of π΄π΅.
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We can then say that πΆ must be here, closer to π΅, as this length of πΆπ΅ would fit.
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Three lots of πΆπ΅ would give us π΄π΅.
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We now need to work out the question of how πΆ divides this vector π΅π΄.
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We can then say if π΅πΆ is one unit length long, then π΄πΆ would be equivalent to two of these lengths.
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So do we write this ratio as two to one or one to two?
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Well, the direction here is very important.
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Weβre given the vector π΅π΄, so that means that weβre going from π΅ to π΄.
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We can therefore give our answer that itβs the ratio one to two, which is given in option (B).
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Note that if we had been given the vector π΄π΅ instead, then that wouldβve been the ratio two to one.
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But here, since πΆ is dividing vector π΅π΄, then itβs the ratio one to two.
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Weβll now look at one final question.
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A bus is traveling from city π΄ at coordinates 10, negative 10 to city π΅ at coordinates negative eight, eight.
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Its first stop is at πΆ, which is halfway between the cities.
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Its second stop is at π·, which is two-thirds of the way from π΄ to π΅.
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What are the coordinates of πΆ and π·?
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In this problem question, we have a bus which is traveling from π΄ to π΅.
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It stops halfway, which is at the point πΆ, and then it stops again at point π·, which is two-thirds of the way from π΄ to π΅.
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We might want to start this by modeling our coordinates on a graph.
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So here we have π΄ at 10, negative 10 and π΅ at negative eight, eight.
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We could even join these with a line.
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The halfway point on this bus journey is at point πΆ.
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If weβd used a graph paper for this and we had a nice integer result, we might be able to read the coordinate of point πΆ directly from the graph.
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But letβs see if we can solve this using the formula to find the midpoint of a line segment.
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This formula tells us for the two points π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two, the midpoint of the line joining these can be found at the coordinate π₯ sub one plus π₯ sub two over two, π¦ sub one plus π¦ sub two over two.
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We can designate the point π΄ with the π₯ sub one, π¦ sub one values and point π΅ with the π₯ sub two and π¦ sub two values, and we then plug these into the formula.
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This gives us that the π₯-value of our midpoint will be 10 plus negative eight over two and the π¦-value will be negative 10 plus eight over two.
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The midpoint can then be written as two over two, negative two over two, which of course is one, negative one.
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We have therefore found our first answer.
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The coordinate of πΆ, which is the midpoint, is at one, negative one.
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Letβs clear some space and see if we can find the coordinates of π·.
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Weβre given that π· is two-thirds of the way from π΄ to π΅.
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We could split our line into three equal parts, and we know that π· is two of these three parts along.
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Although we have a midpoint formula to find a value halfway along a line, we donβt have a formula to find a value two-thirds of the way along the line.
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We might remember, however, that thereβs the section formula, which allows us to split a line segment into a given ratio.
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This tells us that for two points π΄ with coordinates π₯ sub one, π¦ sub one and π΅ with coordinates π₯ sub two, π¦ sub two, the point π, which divides line segment π΄π΅ in the ratio π to π, has coordinates π equal to ππ₯ sub two plus π π₯ sub one over π plus π, ππ¦ sub two plus ππ¦ sub one over π plus π.
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In order to find the coordinates of our point π· which divides this line segment π΄π΅, we need to know the ratio.
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Since we were told that π· is two-thirds of the way from π΄ to π΅, we split π΄π΅ into three parts.
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From π΄ to π·, itβs two out of those three parts, and from π· to π΅, itβs the remaining one part.
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Point π· therefore splits the line segment π΄ to π΅ in the ratio two to one.
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To use this section formula, we need the letters π to π which is the ratio two to one, and we need our two coordinates π΄ and π΅.
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We can therefore plug in our values into the formula to give us that the π₯-value of coordinate π· is two times negative eight plus one times 10 over two plus one.
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And the π¦-value of this coordinate is two times eight plus one times negative 10 over two plus one.
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Simplifying these values gives us that the coordinates of π· are negative six three, six three, which simplifies to negative two, two.
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We can therefore give our two answers πΆ is at the coordinates one, negative one and π· is at the coordinates negative two, two.
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We can now summarize what weβve learned in this video.
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Firstly, we saw the section formula, which tells us that for point π΄ with coordinates π₯ sub one, π¦ sub one and point π΅ with coordinates π₯ sub two, π¦ sub two, the point π which divides line segment π΄π΅ in the ratio π to π has coordinates as follows.
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The π₯-value is π times π₯ sub two plus π times π₯ sub one over π plus π, and the π¦-value is π times π¦ sub two plus π times π¦ sub one over π plus π.
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We also saw that when weβre partitioning or dividing a line into two equal pieces, then we can use the formula to find the midpoint of a line segment.
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The midpoint of the line joining the two coordinates π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two has the coordinates π₯ sub one plus π₯ sub two over two, π¦ sub one plus π¦ sub two over two.
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Finally, a handy tip: itβs very useful to draw the graphs and plot any points that weβre given.
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This will help us think logically through the problems and is useful when weβre finding the ratios in a line segment.