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IB MATHEMATICS AA · TOPIC 2.1 · TOPIC 2 BUNDLE

Lines & Coordinate Geometry

A complete foundation chapter for IB Mathematics AA, now included in the €50 Topic 2 Functions bundle. Learn how gradients describe direction, how equations encode straight lines, why perpendicular gradients multiply to −1, how perpendicular bisectors are constructed, and how exact algebra and GDC methods fit together.

WHAT YOU NEED TO KNOW

Straight lines become easier when every calculation has a geometric meaning.

Students often memorise formulas for gradient and line equations without seeing the picture behind them. This chapter starts with the geometry first. A gradient tells us how quickly a line rises or falls. A midpoint gives the centre of a segment. A perpendicular bisector combines both ideas: first find the midpoint, then turn the direction through 90°. Once those ideas are secure, the algebra becomes much more natural.

01 · GRADIENT

Read direction before calculating

For two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\), gradient compares vertical change with horizontal change. Positive means rising, negative means falling, zero means horizontal, and a vertical line has undefined gradient.

02 · EQUATIONS

Choose the safest form

If you know a point and a gradient, \(y-y_1=m(x-x_1)\) is usually the cleanest start. If you need an intercept quickly, \(y=mx+c\) is often more useful. The notes explain when each form saves time.

03 · PERPENDICULAR

Think 90°, not just “negative reciprocal”

For non-vertical perpendicular lines, \(m_1m_2=-1\). The chapter uses true right-angle sketches so the algebra matches the geometry students are meant to visualise.

04 · INTERSECTIONS

Exact algebra first, technology second

You will solve intersections by substitution or elimination and then see the equivalent graphing route on both the Casio fx-CG50 and TI-Nspire CX II.

MINI WORKED EXAMPLE

A perpendicular line through a known point

Suppose line \(L_1\) passes through \(A(-2,-1)\) and \(B(4,3)\). First calculate its gradient:

\[m_{L_1}=\frac{3-(-1)}{4-(-2)}=\frac{4}{6}=\frac{2}{3}.\]

A perpendicular line therefore has gradient \(-\frac{3}{2}\). If the second line also passes through \(A(-2,-1)\), use point-gradient form:

POINT-GRADIENT FORM\[y+1=-\frac{3}{2}(x+2)\]

The important idea is the order: find the original direction → rotate it by 90° → anchor the new line through the required point.

NEW TOPIC PRICING

Included in Topic 2 · Functions · €50

Lines & Coordinate Geometry is now part of the complete Topic 2 bundle rather than a separate free chapter. The €50 Topic 2 purchase unlocks Lines, Quadratics, Composition & Inverse Functions, Exponentials & Logarithms, and Asymptotes & Rational Functions in My Notes.

Open the complete Topic 2 bundle →

CONTINUE TOPIC 2

Ready for Quadratics?

The next chapter develops graph shape, roots, discriminants, Vieta's formulae, inequalities, optimisation and parameter problems.

Explore Topic 2.2 Quadratics →
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