IP Maths Notes (Lower Sec, Year 1-2): 04) Coordinate Geometry & Graphs

Coordinate geometry links algebra to visuals. This note refreshes gradient logic, distance and midpoint formulas, equation forms, and graph interpretation.

Learning targets

• Compute gradients and intercepts from points or equations.
• Convert between point-slope, slope-intercept, and general linear forms.
• Apply the distance and midpoint formulas accurately.
• Interpret real-world scenarios from line graphs (rate, supply-demand, temperature).

1. Gradient and intercepts

• Gradient (slope) between points \( (x_1, y_1) \) and \( (x_2, y_2) \):

\[ m = \frac{y_2 - y_1}{x_2 - x_1}. \]

• Slope-intercept form: \( y = mx + c \), where \( c \) is the y-intercept.
• Point-slope form: \( y - y_1 = m(x - x_1) \).

Worked example — Gradient & intercept

Find the gradient and y-intercept of the line passing through \( (2, -3) \) and \( (5, 6) \).

\[ m = \frac{6 - (-3)}{5 - 2} = \frac{9}{3} = 3. \]

Use point-slope form with \( (2, -3) \):

\[ y + 3 = 3(x - 2) \implies y = 3x - 9. \]

Y-intercept is \( -9 \).

2. Distance and midpoint formulas

• Distance between two points: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
• Midpoint: \( M\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \).

Example: Points \( A(-2, 5) \) and \( B(4, -1) \).

\[ d = \sqrt{(4 - (-2))^2 + (-1 - 5)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{72} = 6\sqrt{2}. \]

Midpoint:

\[ M\left(\frac{-2 + 4}{2}, \frac{5 + (-1)}{2}\right) = M(1, 2). \]

3. Parallel and perpendicular lines

• Parallel lines share the same gradient.
• Perpendicular gradients multiply to \( -1 \): if \( m_1 \times m_2 = -1 \), the lines are perpendicular.

Example: The line through \( (3, -4) \) perpendicular to \( y = -\frac{1}{2}x + 1 \) has gradient \( 2 \) and equation \( y + 4 = 2(x - 3) \implies y = 2x - 10 \).

4. Graph interpretation

• Always label axes with units.
• Identify intercepts, turning points, and intersection points.
• Connect slope to rate (e.g., distance-time graphs: slope = speed).

Worked example — Interpreting a distance-time graph

A student jogs 3 km in 18 minutes, rests for 6 minutes, then jogs another 2 km in 10 minutes.

• Segment 1 gradient: \( \frac{3}{0.3} = 10 \text{ km h}^{-1} \).
• Segment 2 gradient: 0 (rest).
• Segment 3 gradient: \( \frac{2}{\frac{10}{60}} = 12 \text{ km h}^{-1} \).

Explain what each segment means in sentences — examiners award communication marks.

5. System of equations via intersection

Graphing \( y = 2x + 1 \) and \( y = -x + 7 \) reveals intersection at \( (2, 5) \). Solving algebraically verifies the result.

Try it yourself

1. Find the equation of the line through \( (4, -5) \) with gradient \( -\frac{3}{4} \).
2. Determine the midpoint of the segment joining \( (7, 2) \) and \( (-1, -6) \).
3. Show that the line joining \( (2, 3) \) and \( (6, 7) \) is parallel to the line \( 2x - 2y = 1 \).
4. Two lines intersect at \( (1, -2) \). One has equation \( y = 4x - 6 \). Find the equation of the perpendicular line through the same point.

Advance to quadratics at https://eclatinstitute.sg/blog/ip-maths-lower-sec-notes/IP-Maths-Lower-Sec-05-Quadratic-Expressions-and-Graph-Sketching.