The Three Tools#
Analytical Geometry gives you three formulas. Every question in this section uses one or more of them.
1. The Distance Formula#
Logic: The distance between two points is the hypotenuse of a right-angled triangle (Pythagoras!).
$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$Worked Example#
Find the distance between $A(1; 3)$ and $B(4; 7)$.
$d = \sqrt{(4-1)^2 + (7-3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
2. The Midpoint Formula#
Logic: The midpoint is the average of the coordinates.
$$ M = \left( \frac{x_1 + x_2}{2} ; \frac{y_1 + y_2}{2} \right) $$Worked Example#
Find the midpoint of $A(2; -1)$ and $B(6; 5)$.
$M = \left(\frac{2+6}{2} ; \frac{-1+5}{2}\right) = (4; 2)$
Reverse Midpoint#
If $M(3; 4)$ is the midpoint of $A(1; 2)$ and $B(x; y)$:
$\frac{1 + x}{2} = 3 \Rightarrow x = 5$
$\frac{2 + y}{2} = 4 \Rightarrow y = 6$
So $B = (5; 6)$.
3. The Gradient Formula#
Logic: Gradient is the rate of change — “Rise over Run.”
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$| Gradient | Meaning |
|---|---|
| $m > 0$ | Line goes UP (left to right) |
| $m < 0$ | Line goes DOWN |
| $m = 0$ | Horizontal line |
| Undefined | Vertical line ($x_1 = x_2$) |
Worked Example#
Find the gradient of the line through $(2; 5)$ and $(6; -3)$.
$m = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2$
4. Parallel and Perpendicular Lines#
Parallel Lines#
Two lines are parallel if they have the same gradient: $m_1 = m_2$
Perpendicular Lines#
Two lines are perpendicular if their gradients multiply to $-1$: $m_1 \times m_2 = -1$
Equivalently: $m_2 = -\frac{1}{m_1}$ (the “negative reciprocal”).
Worked Example#
Line AB has gradient $\frac{2}{3}$. Line CD is perpendicular to AB. What is CD’s gradient?
$m_{CD} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}$
5. Collinearity (Points on the Same Line)#
Three points are collinear (on the same straight line) if the gradient between any two pairs is the same.
Worked Example#
Are $A(1; 2)$, $B(3; 6)$, $C(5; 10)$ collinear?
$m_{AB} = \frac{6-2}{3-1} = \frac{4}{2} = 2$
$m_{BC} = \frac{10-6}{5-3} = \frac{4}{2} = 2$
$m_{AB} = m_{BC}$ ✓ → The points are collinear.
6. Proving Properties of Shapes#
Exam questions often give you 3 or 4 points and ask you to prove the shape is a parallelogram, rectangle, rhombus, etc.
| To prove it’s a… | Show that… |
|---|---|
| Parallelogram | Both pairs of opposite sides are parallel ($m_1 = m_2$) |
| Rectangle | It’s a parallelogram AND one angle is $90°$ ($m_1 \times m_2 = -1$) |
| Rhombus | All 4 sides are equal (use distance formula) |
| Square | All sides equal AND one angle is $90°$ |
| Right angle at a vertex | Two sides meeting at that vertex are perpendicular |
Worked Example#
Show that $A(1; 1)$, $B(4; 1)$, $C(4; 5)$ form a right-angled triangle, and identify the right angle.
$m_{AB} = \frac{1-1}{4-1} = \frac{0}{3} = 0$ (horizontal line)
$m_{BC} = \frac{5-1}{4-4} = \frac{4}{0}$ → undefined (vertical line)
A horizontal line is perpendicular to a vertical line, so $\hat{B} = 90°$.
∴ $\triangle ABC$ is right-angled at $B$. ✓
💡 When one gradient is 0 and the other is undefined, the lines are automatically perpendicular — you don’t need to check $m_1 \times m_2 = -1$.
Alternative with non-trivial gradients: Show that $P(0; 0)$, $Q(4; 2)$, $R(3; -6)$ has a right angle.
$m_{PQ} = \frac{2}{4} = \frac{1}{2}$, $\quad m_{PR} = \frac{-6}{3} = -2$
$m_{PQ} \times m_{PR} = \frac{1}{2} \times (-2) = -1$ ✓
∴ $PQ \perp PR$, so $\hat{P} = 90°$.
🚨 Common Mistakes#
- Subtracting coordinates in wrong order: Be consistent — always do $(x_2 - x_1)$ in both numerator and denominator. Swapping the order in one but not the other flips the sign.
- Forgetting to square in distance formula: $d = \sqrt{(3)^2 + (4)^2} = \sqrt{25} = 5$, NOT $\sqrt{3 + 4} = \sqrt{7}$.
- Perpendicular gradient: It’s the NEGATIVE reciprocal. If $m = 2$, perpendicular is $-\frac{1}{2}$, NOT $\frac{1}{2}$.
- Undefined gradient: If $x_1 = x_2$, the gradient is undefined (vertical line). Don’t write $m = 0$.
- Not stating reasons: In proofs, always write WHY (e.g., “AB ∥ CD because $m_{AB} = m_{CD} = 2$”).
💡 Pro Tip: The Equation of a Line#
Once you have the gradient $m$ and a point $(x_1; y_1)$, the equation of the line is:
$$ y - y_1 = m(x - x_1) $$This is the point-gradient form and it’s the fastest way to find any line’s equation.
🔗 Related Grade 10 topics:
- Trig Ratios — gradient = $\tan\theta$, and the distance formula comes from Pythagoras
- Solving Equations — simultaneous equations find where two lines intersect
- Sketching Graphs — the gradient $a$ in $y = ax + q$ IS the gradient from this section
📌 Where this leads in Grade 11:
- Angle of Inclination, Circles & Tangents — angle between lines, equation of a circle, tangent to a circle