Geometry and Trigonometry

The distance formula is just Pythagoras rotated onto a coordinate grid: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$, and in 3D you simply add a $(z_2-z_1{)}^2$ term for problems like an aircraft's position given as $(x,y,z)$ in km. Midpoint is an average, not a difference — it's $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$, and mixing that up under pressure is a frequent slip. Both formulas are formula-booklet safe, so the real skill being tested is reading the coordinates correctly out of a diagram or word problem.

Examiners test the perpendicular bisector of $[AB]$ as the set of ALL points equidistant from $A$ and $B$ — not just 'a line at 90° through the midpoint.' To build it: find the midpoint of $AB$, find the gradient of $AB$, take the negative reciprocal for the perpendicular gradient ($m_1{m}_2=-1$), then substitute the midpoint into $y-y_1=m_{\perp }(x-x_1)$. If a triangle is isosceles with $AC=BC$, point $C$ automatically lies on this bisector — a one-line 'hence' argument that scores well without rebuilding the whole line equation.

The shoelace formula $\text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ gives a triangle's area directly from three vertices, but it is NOT in the formula booklet, so memorise it or fall back on the cosine-rule route. At HL, the same result comes from the 2D vector cross product, $\text{Area}=\frac{1}{2}\left|x_{AB}{y}_{AC}-x_{AC}{y}_{AB}\right|$, tested alongside the Vectors topic. Running Area $=\frac{1}{2}$ base × height BACKWARDS is also a clean way to find a perpendicular distance from a point to a line once the area and base length are known.

SOHCAHTOA still runs right-angled triangle problems: $\sin \theta =\frac{\text{opp}}{\text{hyp}}$, $\cos \theta =\frac{\text{adj}}{\text{hyp}}$, $\tan \theta =\frac{\text{opp}}{\text{adj}}$, but a genuine right angle must exist first. Angle of elevation is measured UP from the horizontal, angle of depression DOWN from it, and the two are equal by alternate angles when the horizontal lines are parallel — which is why a 'looking down from a rooftop' problem is the same triangle as 'looking up from the ground'. Pythagoras is the silent partner here, usually needed to find a missing side before an angle can even be calculated.

For the sine and cosine rules to work on real, often obtuse, triangles, $\sin \theta$ and $\cos \theta$ must be defined beyond $0°$–$90°$: on the unit circle, $\cos \theta$ is the $x$-coordinate and $\sin \theta$ is the $y$-coordinate of the point at angle $\theta$ anticlockwise from the positive $x$-axis. The key consequence: $\sin (180°-\theta )=\sin \theta$ but $\cos (180°-\theta )=-\cos \theta$ — sine stays positive in the second quadrant, cosine flips sign, which is exactly why $\arccos$ can return a negative value and still describe a valid obtuse angle. The identity ${\sin }^2\theta +{\cos }^2\theta =1$ falls straight out of this circle, and radians ($\pi$ radians $=180°$) show up increasingly often at HL.