Year 10 Coordinate Geometry Worksheets

Worked examples

Year 10 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 · Parallel and perpendicular gradients

   Line $\ell$ has equation $y = 3x - 5$. Write down the gradient of any line that is (a) parallel to $\ell$, and (b) perpendicular to $\ell$.

   Show worked solution

   Key idea: Parallel lines have equal gradients. Perpendicular lines have gradients whose product is $-1$, so the perpendicular gradient is the negative reciprocal.

   1. The gradient of $\ell$ is $m = 3$.
   2. (a) Parallel gradient: A parallel line has the same gradient, so $m_{\parallel} = 3$.
   3. (b) Perpendicular gradient: Flip and negate: $m_{\perp} = -\dfrac{1}{3}$.
   4. Check: $3 \times \left(-\dfrac{1}{3}\right) = -1$ ✓

   Answer: Parallel gradient $= 3$; perpendicular gradient $= -\dfrac{1}{3}$.

2. Example 2 · Equation of a perpendicular line through a point

   Find the equation of the line that passes through the point $(2, -1)$ and is perpendicular to the line $2x - 4y + 3 = 0$. Write your answer in the form $y = mx + c$.

   Show worked solution

   Key idea: Rearrange the given line to find its gradient, then use the negative reciprocal for the perpendicular gradient, and substitute the point into $y - y_1 = m(x - x_1)$.

   1. Rearrange $2x - 4y + 3 = 0$: $\;4y = 2x + 3 \Rightarrow y = \dfrac{1}{2}x + \dfrac{3}{4}$, so $m = \dfrac{1}{2}$.
   2. Perpendicular gradient: $m_{\perp} = -\dfrac{1}{\frac{1}{2}} = -2$.
   3. Substitute $(2,\,-1)$ and $m = -2$ into $y - y_1 = m(x - x_1)$:
      $y - (-1) = -2(x - 2)$
   4. Expand and simplify:
      $\begin{aligned} y + 1 &= -2x + 4 \\ y &= -2x + 3 \end{aligned}$
   5. Check: Substitute $(2, -1)$: $-2(2) + 3 = -1$ ✓

   Answer: $y = -2x + 3$

3. Example 3 · Real-life perpendicular path problem

   A straight walking track runs through the points $A(1, 3)$ and $B(5, 7)$, where units are in kilometres. A ranger station is located at $C(6, 2)$. A supply path is to be built from $C$ to the nearest point on track $AB$, meeting the track at a right angle. Find the equation of the supply path, and find the coordinates of the point where it meets the track.

   Show worked solution

   Key idea: The supply path is perpendicular to $AB$ and passes through $C$. Its equation combined with the equation of $AB$ gives the meeting point.

   1. Gradient of $AB$: $m_{AB} = \dfrac{7 - 3}{5 - 1} = \dfrac{4}{4} = 1$.
   2. Perpendicular gradient: $m_{\perp} = -1$.
   3. Equation of supply path through $C(6, 2)$:
      $\begin{aligned} y - 2 &= -1(x - 6) \\ y &= -x + 8 \end{aligned}$
   4. Equation of track $AB$ through $A(1,3)$ with $m = 1$: $y = x + 2$.
   5. Solve simultaneously to find the meeting point:
      $\begin{aligned} -x + 8 &= x + 2 \\ 6 &= 2x \\ x &= 3, y = 5 \end{aligned}$
   6. Check: Point $(3, 5)$: on $AB$: $3 + 2 = 5$ ✓; on supply path: $-3 + 8 = 5$ ✓

   Answer: Supply path equation: $y = -x + 8$; meeting point: $(3, 5)$.

Common mistakes

Where students usually lose marks

Try one yourself

Your turn

Got a different coordinate geometry question wrong? Upload it to Tuterly and get the worked solution and a fresh one to practise, free.