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Finding the co-ordinates of collinear points

You may also be asked to find the co-ordinates of a point which divides the directed line segment in a given ratio. There are many different ways of writing this down. The model we're showing you below is one of the simpler ways!

Find the co-ordinates of P which divides A(2, 7, 8) and B(7, 17, 28) in the ratio 3:2.

Point P dividing line

Point P on line


Choose two directed line segments. One which does not involve P and one which does involve P.

\eqalign{ \overrightarrow {AB} & = & \left( \matrix{ 5 \cr 10 \cr 20 \cr} \right) \cr \overrightarrow {AP} & = & ?\cr}

Write down the ratio for the chosen line segments

\overrightarrow {AP} :\;\overrightarrow {AB} = 3:\;5

i.e. {{\overrightarrow {AP} } \over {\overrightarrow {AB} }} = {3 \over 5}

so \overrightarrow {AP} = {3 \over 5}\overrightarrow {AB}

Find the collinear vector

\overrightarrow {AP} = {3 \over 5}\;\left( \matrix{ 5 \cr 10 \cr 20 \cr} \right) = \left( \matrix{ 3 \cr 6 \cr 12 \cr} \right)

Complete your working

\overrightarrow {AP} starts at A=(2, 7, 8), finishes {\rm{P = (5, 13, 20)}}


Find the co-ordinates of T which divides A(1, -2, 7) and B(9, 10, -5) in the ratio 1:3.

Point T dividing line

Point T dividing line



\overrightarrow {AB} = \left( \matrix{ 8 \cr 12 \cr - 12 \cr} \right)

\overrightarrow {AT} = ...

\overrightarrow {AT} :\;\overrightarrow {AB} = 1\;:\;4 i.e. {{\overrightarrow {AT} } \over {\overrightarrow {AB} }} = {1 \over 4}

so \overrightarrow {AT} = {1 \over 4}\;\overrightarrow {AB}

\overrightarrow {AT} = {1 \over 4}\;\left( \matrix{ 8 \cr 12 \cr - 12 \cr} \right) = \left( \matrix{ 2 \cr 3 \cr - 3 \cr} \right)

\overrightarrow {AT} starts at A=(1, -2, 7),finishes at T=(3, 1, 4)

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