A straight-line equation is a compact description of constant rate of change. In Additional Mathematics, line work becomes stronger when algebraic forms, graph features and geometric relationships are treated as the same object.
1. Gradient measures rate of change
For two distinct points (x₁,y₁) and (x₂,y₂), the gradient is m=(y₂−y₁)/(x₂−x₁), provided x₂≠x₁.
A positive gradient rises from left to right; a negative gradient falls; zero gives a horizontal line. A vertical line has undefined gradient because its horizontal change is zero.
2. y=mx+c reveals gradient and vertical intercept
In y=mx+c, m is the gradient and c is the y-coordinate where x=0. For y=3x−5, the gradient is 3 and the vertical intercept is −5.
3. Point-gradient form is often faster
A line of gradient m through (x₁,y₁) can be written y−y₁=m(x−x₁).
Example: a line of gradient −2 through (3,5) is y−5=−2(x−3), or y=−2x+11.
4. Find a line through two points
Through A(1,4) and B(5,12), m=(12−4)/(5−1)=2. Using point-gradient form with A gives y−4=2(x−1), hence y=2x+2.
Check with B: 12=10+2.
5. Horizontal and vertical intercepts
To find the vertical intercept, set x=0. To find the horizontal intercept, set y=0.
For 2x+3y=12, setting x=0 gives y=4; setting y=0 gives x=6. The line passes through (0,4) and (6,0).
6. General form can hide useful structure
For Ax+By+C=0 with B≠0, rearrange to y=−(A/B)x−C/B. The gradient is −A/B.
Thus 4x−2y+7=0 has gradient 2 because y=2x+7/2.
7. Intersection of two lines
For y=2x+1 and y=−x+7, equate outputs: 2x+1=−x+7. Then 3x=6, so x=2 and y=5. The intersection is (2,5).
Geometrically, the point satisfies both line equations simultaneously.
8. Gradient as a geometric test
Equal gradients indicate parallel nonvertical lines. For perpendicular nonvertical lines with finite nonzero gradients m₁ and m₂, m₁m₂=−1.
Special cases matter: horizontal and vertical lines are perpendicular even though the vertical gradient is undefined.
9. Common mistakes
- Reversing only one difference in the gradient formula.
- Calling c the horizontal intercept.
- Using m₁m₂=−1 when one line is vertical.
- Finding an equation but not checking the supplied point.
10. Practice
- Find the gradient through (2,3) and (6,15).
- Find the equation of the line of gradient 4 through (1,−2).
- Find both intercepts of 3x+2y=12.
- Find the intersection of y=3x−1 and y=x+5.
- State the gradient of 6x+3y−9=0.
11. Answers
- 3.
- y+2=4(x−1), so y=4x−6.
- (4,0) and (0,6).
- 3x−1=x+5 gives x=3, y=8.
- −2.
Continue with the Coordinate Geometry guide, Linear Law, or return to the Additional Mathematics Hub.