, as you scroll — highermaths.org highermaths.org read instead ›

Straight lines

Two points are all a line needs.

Pick any two points and exactly one straight line runs through both. How steep it is, where it crosses, what meets it at right angles — all of it comes from those four coordinates.

scroll ↓

Gradient

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

From AA to BB you go 88 across and 66 up, so m=68=34m = \tfrac{6}{8} = \tfrac{3}{4}: every step right climbs three-quarters of a step.

Midpoint

(x1+x22,  y1+y22)\left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2}\right)

Halfway between two points is the average of their coordinates: (−3+52, 2+82)=(1, 5)\left(\tfrac{-3 + 5}{2},\, \tfrac{2 + 8}{2}\right) = (1,\,5).

Distance

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The length ABAB is the hypotenuse of the very triangle that gave the gradient: 82+62=100=10\sqrt{8^2 + 6^2} = \sqrt{100} = 10. Distance is just Pythagoras.

Perpendicular

m1m2=−1m_1 m_2 = -1

Turn the line through 90∘90^\circ and its triangle turns with it: 8 across and 6 up becomes 6 across and 8 down. So the gradient flips and changes sign — 34\tfrac{3}{4} becomes −43-\tfrac{4}{3}, and 34×−43=−1\tfrac{3}{4} \times -\tfrac{4}{3} = -1.

Chapter 4

One triangle did all of it.

Parallel lines, the ax+by+c=0ax + by + c = 0 form, modelling with straight lines, and twelve graded exam questions with full worked solutions are waiting.

Read the full chapter › Jump to the questions ›