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To find the midpoint of a line between two sets of coordinates, find the average of the \(x\)-axis coordinates and the average of the \(y\)-axis coordinates.

Example

Find the midpoint of the line joining the point D \((10, 5)\) and E \((4, 11)\).

Average of \(x\)-coordinates       \(\frac{10+4}{2}=7\)

Average of \(y\)-coordinates       \(\frac{5+11}{2}=8\)

The midpoint is \((7, 8)\).

It may be helpful to set out your work like this:

Example

P is the point with coordinates \((–5, 5)\) and Q is the point with coordinates \((x, y)\).

M is the midpoint of PQ and has coordinates \((–1, 3)\).

Find the coordinates of Q.

Solution

In this question, the midpoint is known and one of the end points has to be found.

Use \((x, y)\) for the coordinates of Q and apply the same method as before.

Set up an equation using \(x\)-coordinates \(\frac{x–5}{2}–1\) M is the point \((–1, 3)\)

Solve equation

• Multiply across by 2 ��(��–5=–2��)
  
  
• Add 5 to each side ��(��–5+5=–2+5��)
  \(x=3\)

Set up an equation using \(y\)-coordinates

\(\frac{5+y}{2}=3\)M is the point \((–1, 3)\)

Solve equation \(5+y=6\)

\(y=1\)
Coordinates of Q are \((3,1)\)

Answer

The coordinates of Q are \((3,1)\)

To find the length of a line between points

• Subtract the \(x\)-coordinates
• Subtract the \(y\)-coordinates
• Apply Pythagoras' theorem to these values

Example

Find the length of the line joining D \((6, 9)\) and E \((–2, 3)\)

Solution

• Subtract the \(x\)-coordinates

��(6–2=6+2=8��)

• Subtract the \(y\)-coordinates

��(9–3=6��)

• Apply Pythagoras' theorem to these values

\(length=\sqrt{8^2+6^2}\)

\(=\sqrt{100}\)

\(=10\)

It will be helpful to set out your work like this:

\(\matrix{ & &~x & ~y \cr Coords~of~D & &~6 & ~9 \cr Coords~of~E & - & -2 & ~3 \cr Subtract & & \hline ~8 & ~6}\)

Pythagoras Length = \(\sqrt{8^2 + 6^2}\)    = \(\sqrt{100}\)    = \(10\)