Title: InkMaths / Coordinate geometry and vectors
On every note of this deck, unless a note says otherwise:
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Front: STANDARD RESULT — LEARN IT
Circle from the expanded form:
x^2 + y^2 + 2gx + 2fy + c = 0: centre? radius?\(x^{2} + y^{2} + 2gx + 2fy + c = 0: \text{centre?} \text{radius?}\)x^2 + y^2 + 2gx + 2fy + c = 0: centre? radius?
Back: (-g, -f), √g^2 + f^2 - c\((-g, -f), \sqrt{g^{2} + f^{2} - c}\)(-g, - f), √(g^2 + f^2 - c)

Front: STANDARD RESULT — LEARN IT
Magnitude and unit vector:
a = xi + yj + zk: |a| = ?, unit vector = ?\(\text{a} = x\text{i} + y\text{j} + z\text{k}: |\text{a}| = ?, \text{unit vector} = ?\)a = x i + y j + z k: | a| = ?, unit vector = ?
Back: √x^2 + y^2 + z^2, a|a|\(\sqrt{x^{2} + y^{2} + z^{2}}, \frac{\text{a}}{|\text{a}|}\)√(x^2 + y^2 + z^2), (a/| a|)

Front: RECALL
State how to turn parametric equations into a cartesian one.
Back: Make the parameter the subject of whichever equation is easier and substitute it into the other. With trig parameters use cos²t + sin²t = 1 instead of solving for t. Then say what range of x and y the parameter allows, because the cartesian equation on its own describes more curve than you have.

Front: SPOT THE ERROR
Say what is wrong with this claim:
If A and B have position vectors a and b, the vector from A to B is a + b.
Back: It is b − a. Travel back down a to the origin, then out along b. Adding the two position vectors gives you the diagonal of a parallelogram, which is a different point entirely, and the midpoint of AB is ½(a + b) rather than a + b.

Front: STANDARD RESULT — LEARN IT
Perpendicular gradient (m ≠ 0):
a line of gradient m: perpendicular gradient = ?\(\text{a line of gradient m}: \text{perpendicular gradient} = ?\)a line of gradient m: perpendicular gradient = ?
Back: -1m, because m_1m_2 = -1. A horizontal line's perpendicular is vertical, with no gradient\(-\frac{1}{m}, \text{ because } m_{1}m_{2} = -1. \text{ A horizontal line's perpendicular is vertical, with no gradient}\)-(1/m), because m_1m_2 = -1. A horizontal line's perpendicular is vertical, with no gradient

Front: RECALL
State the three circle facts a coordinate question leans on.
Back: The angle in a semicircle is a right angle. The perpendicular from the centre bisects a chord. A tangent meets the radius at the point of contact at a right angle. Almost every circle question is one of these three plus the distance formula.