Coordinate Geometry and Graphs — NSW Year 9 Advanced Maths
Distance, midpoint, gradient, straight lines, direct proportion, parabolas and circles. Twelve subtopics and 240 questions with worked solutions, aligned to NESA outcomes MA5-LIN-C-01, MA5-LIN-C-02 and MA5-NLI-P-01.
Coordinate Geometry and Graphs in NSW Year 9 Advanced
Coordinate geometry links algebra with graphs: every equation draws a picture, and every graph has an equation. In Year 9 Advanced, students find the length, midpoint and gradient of an interval, graph straight lines and find their equations, solve equations graphically, model direct proportion and real situations with lines, and graph parabolas and circles.
The twelve subtopics below move from the length of an interval, midpoints and gradient to graphing lines, the gradient-intercept equation, finding a line's equation and solving equations graphically, then direct proportion and applying linear equations, and finish with parabolas and circles. Coordinate Geometry and Graphs is one of the 14 topics in NSW Year 9 Advanced Maths and belongs to NESA's Number and Algebra strand (NESA outcomes MA5‑LIN‑C‑01, MA5‑LIN‑C‑02, MA5‑LIN‑P‑01, MA5‑RAT‑P‑01, MA5‑NLI‑C‑02 and MA5‑NLI‑P‑01). It leads on to simultaneous equations and further curves.
Subtopics
Coordinate Geometry and Graphs: common questions
How do you find the distance between two points?
Use \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\), which comes from Pythagoras' theorem. From \(A(3, 4)\) to \(B(8, 16)\), \(d = \sqrt{5^2 + 12^2} = 13\) units.
What do m and c mean in y = mx + c?
In \(y = mx + c\), \(m\) is the gradient and \(c\) is the \(y\)-intercept. The line \(y = 5 - 2x\) has gradient \(-2\) and crosses the \(y\)-axis at 5.
How do you find a line's equation from two points?
Find the gradient, then substitute one point into \(y = mx + c\) to find \(c\). Through \((1, 3)\) and \((2, 5)\) the gradient is 2, and \(3 = 2 + c\) gives \(c = 1\), so \(y = 2x + 1\).
How do you find the vertex of a parabola?
For \(y = ax^2 + bx + c\), the axis of symmetry is \(x = -\frac{b}{2a}\), and substituting that \(x\)-value gives the vertex. For \(y = x^2 + 6x + 5\), \(x = -3\) and the vertex is \((-3, -4)\).
Which NESA outcomes does this topic cover?
Distance, midpoint, gradient and graphing lines are MA5‑LIN‑C‑01, and \(y = mx + c\) is MA5‑LIN‑C‑02 (Stage 5 Core). Lines in general form use MA5‑LIN‑P‑01, direct proportion MA5‑RAT‑P‑01, and parabolas and circles MA5‑NLI‑P‑01, with problems about parabolas in MA5‑NLI‑C‑02.