Chapter 5

Straight-Line Graphs

Gradient, midpoint and length, the perpendicular condition, and the standard 0606 question that asks you to find a fourth vertex or the area of a shape from nothing but coordinates.

หลักสูตรแกนกลาง / สสวท.: จำนวนและพีชคณิต (Number and Algebra)ตรงกับ ม.4 เรขาคณิตวิเคราะห์ (เพิ่มเติม) เรื่องเส้นตรง
Common Core (US): G-GPE, F-IF
Cambridge: IGCSE 0606 — Topic 7: Straight-line graphs

5.1 Gradient, midpoint and length

The three that everything else is built from

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}) d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

The distance formula is Pythagoras with the right-angled triangle left undrawn.

Parallel and perpendicular

parallel: m_{1} = m_{2} perpendicular: m_{1} m_{2} = - 1

Perpendicular gradients are negative reciprocals: turn the fraction over and change the sign. A horizontal line and a vertical one are the exception, since a vertical gradient does not exist.

ตัวอย่าง

Find the equation of the perpendicular bisector of the line joining $A (- 1, 2)$ and $B (5, 6)$ .

วิธีทำ

1.Midpoint: $(\frac{- 1 + 5}{2}, \frac{2 + 6}{2}) = (2, 4)$
2.Gradient of $A B$ : $\frac{6 - 2}{5 - ( - 1 )} = \frac{4}{6} = \frac{2}{3}$
3.Perpendicular gradient: $- \frac{3}{2}$
4. $y - 4 = - \frac{3}{2} (x - 2)$

ตอบ $2 y + 3 x = 14$

เคล็ดลับ

A perpendicular bisector needs two ingredients and students usually remember only one. It passes through the MIDPOINT and it has the PERPENDICULAR gradient — the word "bisector" supplies the first and the word "perpendicular" the second.

📖 2 more sections of this chapter are being checked

The whole chapter is written; we read it back section by section before releasing it, because maths nobody has checked should not be on a public page. What is open is complete — nothing here is abridged.

Read it — now practise it

Generate a worksheet on this topic, or sit a timed mock paper.

Build a worksheet Practice arena