IP AMaths Notes (Upper Sec, Year 3-4): 06) Coordinate Geometry of Lines

Linear coordinate geometry blends algebra with spatial reasoning. Track gradients and midpoints carefully and always label key coordinates on a sketch.

Essentials

• Gradient between \( (x_1, y_1) \) and \( (x_2, y_2) \): \( m = \dfrac{y_2 - y_1}{x_2 - x_1} \).
• Distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
• Midpoint: \( \biggl(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\biggr) \).
• Perpendicular lines: \( m_1 m_2 = -1 \) (for non-vertical lines).
• Equation forms: point-slope \( y - y_1 = m(x - x_1) \), two-point, intercept, and normal forms are interchangeable.

Worked example 1 — Perpendicular bisector

Find the equation of the perpendicular bisector of segment joining \( A(2, -1) \) and \( B(8, 5) \).

1. Midpoint: \( M = \left(\dfrac{2 + 8}{2}, \dfrac{-1 + 5}{2}\right) = (5, 2) \).
2. Gradient of \( AB \): \( m_{AB} = \dfrac{5 - (-1)}{8 - 2} = \dfrac{6}{6} = 1 \).
3. Perpendicular gradient: \( m_\perp = -1 \).
4. Equation through \( M \): \( y - 2 = -1(x - 5) \) so \( y = -x + 7 \).

Worked example 2 — Distance from point to line

Compute the distance from \( P(4, -3) \) to the line \( 3x - 4y - 20 = 0 \).

1. Use formula \( \text{dist} = \dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^{2} + B^{2}}} \).
2. Here \( A = 3 \), \( B = -4 \), \( C = -20 \), \( (x_0, y_0) = (4, -3) \).
3. Substitute: numerator \( |3 \times 4 + (-4)(-3) - 20| = |12 + 12 - 20| = |4| = 4 \).
4. Denominator: \( \sqrt{3^{2} + (-4)^{2}} = \sqrt{9 + 16} = 5 \).
5. Distance: \( d = \dfrac{4}{5} \).

Try this

Line \( L_1 \) passes through \( (1, 7) \) and has gradient \( -\tfrac{3}{2} \). Another line \( L_2 \) passes through \( (k, -2) \) and is perpendicular to \( L_1 \). Determine \( k \) if the intersection lies on the \( x \)-axis.