Straight-Line Graphs

Coordinate geometry of the straight line is 0606’s toolkit topic: rarely glamorous, always present, and feeding directly into circle geometry, linear-form reduction and tangent/normal questions in calculus. Every formula here must be instant.

The four basic tools

For $A(x_1, y_1)$, $B(x_2, y_2)$:

• Gradient: $m = \frac{y_2 - y_1}{x_2 - x_1}$, keep the subtraction order consistent top and bottom
• Midpoint: $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
• Length: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, leave as a simplified surd on Paper 1
• Equation of a line: $y - y_1 = m(x - x_1)$, the form that takes a point and gradient directly

Parallel lines: equal gradients. Perpendicular: $m_1 m_2 = -1$, the negative reciprocal. The statement “gradient of perpendicular $= -\frac{1}{m}$” written before use is a method mark in nearly every question that needs it.

The perpendicular bisector routine

The most-asked composite in this topic. For segment AB: midpoint (B/M mark) → gradient of AB, then negative reciprocal (M) → line through the midpoint with that gradient (A). Three visible ingredients, three marking points. Perpendicular bisectors then star in circle questions, the centre lies on the perpendicular bisector of any chord, so the routine pays double.

Area of rectilinear figures: the shoelace

For vertices listed in order (say anticlockwise). $A(1, 2)$, $B(5, 3)$, $C(4, 7)$:

$\text{Area} = \tfrac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$ $= \tfrac{1}{2}|1(3-7) + 5(7-2) + 4(2-3)| = \tfrac{1}{2}|-4 + 25 - 4| = \mathbf{8.5}$

Or the array layout (repeat the first vertex at the end, sum the down-products minus the up-products, halve the absolute value). Two non-negotiables: vertices in order around the shape (a crossed ordering silently halves or garbles the area), and the modulus, area is positive. Writing the array itself earns the method mark even with one arithmetic slip.

Converting to linear form

Relationships like $y = ax^2 + b$ become linear by substitution: plot $y$ against $x^2 \to$ gradient $a$, intercept $b$. The general skill: rewrite the model as $(\text{something}) = m \cdot (\text{something else}) + c$ and say what is plotted against what. The log version ($y = ax^n$, $y = Ab^x$) is the heavyweight case, covered fully in logs & exponentials. When given a table of data: transform the values, state the new variables, and extract constants from gradient and intercept, the statement of what’s plotted is itself a mark.

Worked exam-style question

$A(1, 2)$ and $B(7, 6)$. The perpendicular bisector of $AB$ meets the $x$-axis at $P$. Find the coordinates of $P$.

Midpoint of $AB = (4, 4)$ (B) Gradient $AB = \frac{6-2}{7-1} = \frac{2}{3} \to$ perpendicular gradient $= -\frac{3}{2}$ (M, negative reciprocal shown) Bisector: $y - 4 = -\frac{3}{2}(x - 4)$ (M) At the $x$-axis $y = 0$: $-4 = -\frac{3}{2}(x - 4) \to x - 4 = \frac{8}{3} \to x = \frac{20}{3} \to \mathbf{P\left(\frac{20}{3}, 0\right)}$ (A)

Exact fraction kept to the end, converting to 6.67 mid-stream is how accuracy marks leak.

Common mistakes in this topic

• Gradient subtraction order mixed between numerator and denominator
• Perpendicular gradient as the reciprocal without the sign flip (or vice versa)
• Shoelace applied to unordered vertices
• “Show that $ABC$ is a right angle” answered with lengths when gradients ($m_1 m_2 = -1$) are faster, or Pythagoras attempted with unsimplified surds
• Final form ignored: question asks $ax + by = c$, answer left as $y = mx + c$

Everything here is on the formula list and drilled in week 4 of the revision plan.

Coordinate geometry should be your banker topic. If it’s leaking marks instead, the free 1-hour trial class will find where, message Teacher Rig on WhatsApp.