Year 9 Coordinate Geometry Worksheets

Worked examples

Year 9 coordinate geometry: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Gradient, midpoint and distance between two points

   Two points are $A(1, 2)$ and $B(5, 10)$. Find: (a) the gradient of $AB$, (b) the midpoint of $AB$, and (c) the distance $AB$.

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   Key idea: Use the gradient formula $m = \dfrac{y_2 - y_1}{x_2 - x_1}$, the midpoint formula $M = \left(\dfrac{x_1+x_2}{2},\, \dfrac{y_1+y_2}{2}\right)$, and the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

   1. Gradient: $m = \frac{10 - 2}{5 - 1} = \frac{8}{4} = 2$
   2. Midpoint: $M = \left(\frac{1+5}{2},\, \frac{2+10}{2}\right) = (3,\, 6)$
   3. Distance: $d = \sqrt{(5-1)^2 + (10-2)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$

   Answer: Gradient $= 2$, midpoint $= (3, 6)$, distance $= 4\sqrt{5}$ units.

2. Example 2 · Identifying parallel and perpendicular lines

   Line $\ell_1$ passes through $P(-2, 1)$ and $Q(4, 4)$. Line $\ell_2$ passes through $R(0, 5)$ and $S(6, -7)$. Determine whether $\ell_1$ and $\ell_2$ are parallel, perpendicular, or neither.

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   Key idea: Two lines are parallel if their gradients are equal, and perpendicular if the product of their gradients is $-1$.

   1. Gradient of $\ell_1$: $m_1 = \frac{4 - 1}{4 - (-2)} = \frac{3}{6} = \frac{1}{2}$
   2. Gradient of $\ell_2$: $m_2 = \frac{-7 - 5}{6 - 0} = \frac{-12}{6} = -2$
   3. Check parallel: $m_1 \neq m_2$, so not parallel.
   4. Check perpendicular: $m_1 \times m_2 = \frac{1}{2} \times (-2) = -1 \checkmark$

   Answer: $\ell_1$ and $\ell_2$ are perpendicular.

3. Example 3 · Multi-step coordinate geometry problem

   A park has two water fountains located at $F(1, 3)$ and $G(7, 11)$ on a map, where each unit represents 10 metres. A garden bed is to be planted exactly halfway between the two fountains. A path also runs through the points $H(0, 6)$ and $K(3, 0)$. (a) Find the coordinates of the garden bed. (b) Find the actual straight-line distance between the two fountains in metres. (c) Determine the gradient of the path $HK$ and explain what it means in context.

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   Key idea: Apply the midpoint, distance and gradient formulas, then interpret the answers using the map scale.

   1. Midpoint (garden bed): $M = \left(\frac{1+7}{2},\, \frac{3+11}{2}\right) = (4,\, 7)$
   2. Distance $FG$ in map units: $d = \sqrt{(7-1)^2 + (11-3)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ units}$
   3. Actual distance: $10 \times 10 = 100$ metres.
   4. Gradient of path $HK$: $m = \frac{0 - 6}{3 - 0} = \frac{-6}{3} = -2$
   5. Interpretation: A gradient of $-2$ means the path falls 2 units on the map for every 1 unit across, so it runs steeply downward from left to right.

   Answer: Garden bed at $(4, 7)$; distance between fountains is 100 metres; gradient of path $HK$ is $-2$ (steeply downward from left to right).

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