Algebra · Straight-line graphs

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Finding the equation of a line

How gradient + one point pins down a unique line, why c comes from substituting the point, and why disguised forms like 2y = 6x + 4 must be rearranged before reading anything off.

Algebra · Straight-line graphs

Finding the equation of a line

How gradient + one point pins down a unique line, why c comes from substituting the point, and why disguised forms like 2y = 6x + 4 must be rearranged before reading anything off.

Why it works

A line is pinned down by two facts — a gradient and a point, or two points. Finding "the equation of the line" means recovering

y = mx + c

from whichever two facts you're given.

Gradient + point: substitute to find $c$ . Gradient

-3

through

(2, 5)

: start

y = -3x + c

, and force the line through the point —

(2, 5)

must satisfy the equation:

5 = -3(2) + c \;\Rightarrow\; c = 11 \;\Rightarrow\; y = -3x + 11.

The point's

x

goes with

x

, its

y

with

y

— and the final equation should be CHECKED by substituting the point back in.

Two points: gradient first, then $c$ . Through

(1, 5)

and

(3, 11)

m = \frac{11 - 5}{3 - 1} = 3

, then use either point:

5 = 3(1) + c

gives

c = 2

, so

y = 3x + 2

. Check with the OTHER point:

3(3) + 2 = 11

. ✓ (A graph showing two marked points is the same question in a picture.)

Disguised forms must be undressed first. The gradient of

2y = 6x + 4

is NOT 6 — divide through:

y = 3x + 2

, gradient 3. Similarly

x + 2y = 8

rearranges to

y = 4 - \frac{x}{2}

: gradient

-\frac{1}{2}

, intercept 4. Reading

m

c

off any form that isn't exactly

y = mx + c

is the single biggest error in this topic — one rearrangement first, always.

From a graph: read two GOOD points. Choose lattice points far apart (closer points magnify reading errors), compute

m

, read

c

where the line crosses the

y

-axis — or substitute if the intercept is off the grid.