3 Coordinate Geometry
3.1 Distance, midpoint and slope
For points \(A(x_1, y_1)\) and \(B(x_2, y_2)\):
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, \qquad M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right), \]
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_1 \neq x_2). \]
Two non-vertical lines are parallel when \(m_1 = m_2\), and perpendicular when \(m_1 m_2 = -1\).
3.2 Equations of straight lines
| Form | Equation | Use when you know |
|---|---|---|
| Slope–intercept | \(y = mx + c\) | Slope and \(y\)-intercept |
| Point–slope | \(y - y_1 = m(x - x_1)\) | A point and the slope |
| General | \(Ax + By + C = 0\) | Any line, including vertical |
Example 3.1 (Perpendicular bisector) Find the equation of the perpendicular bisector of the segment joining \(A(1, 2)\) and \(B(5, -6)\).
Solution. The midpoint is \(M = (3, -2)\) and
\[ m_{AB} = \frac{-6 - 2}{5 - 1} = -2, \]
so the perpendicular bisector has slope \(\tfrac{1}{2}\). Using point–slope form,
\[ y + 2 = \tfrac{1}{2}(x - 3) \quad \Longrightarrow \quad x - 2y - 7 = 0. \]
3.3 Circles
The circle with centre \((a, b)\) and radius \(r\) has equation
\[ (x - a)^2 + (y - b)^2 = r^2, \tag{3.1}\]
which expands to the general form
\[ x^2 + y^2 + Dx + Ey + F = 0, \tag{3.2}\]
with centre \(\left(-\tfrac{D}{2}, -\tfrac{E}{2}\right)\) and radius \(\sqrt{\tfrac{D^2 + E^2}{4} - F}\).
Equation 3.2 only represents a real circle when \(\tfrac{D^2 + E^2}{4} - F > 0\). If the expression is zero the equation describes a single point; if it is negative there are no real points at all.
3.4 Exercises
- Find the distance between \((-3, 4)\) and \((2, -8)\).
- Find the centre and radius of \(x^2 + y^2 - 6x + 8y - 11 = 0\).
- The line \(2x - 3y + 6 = 0\) passes through \(P(a, 4)\). Find \(a\).
Answers
- \(13\).
- Centre \((3, -4)\), radius \(6\).
- \(a = 3\).