IB Maths AI HL Linear Functions & Graphs Paper 1 & 2 Gradient, intercepts, forms ~7 min read

Equations of a Straight Line

Every straight line is fixed by a gradient and one point — or equivalently by two points. From there it slots into one of three standard forms: gradient-intercept, point-gradient, or general. The trick is picking the right starting form and rearranging cleanly.

📘 What you need to know

Gradient between two points: m = (y₂ − y₁) / (x₂ − x₁). In the formula booklet.
Gradient-intercept form: y = mx + c. Reads off the gradient and the y-intercept (0, c) directly.
Point-gradient form: y − y₁ = m(x − x₁). The natural starting form when you have a point and a gradient.
General form: ax + by + d = 0. x-intercept (−d/a, 0); y-intercept (0, −d/b).
To get an equation from two points: find m, drop into point-gradient form, then rearrange to the requested form.
GDC shortcut: enter the two points in stats mode and read off the linear regression y = ax + b.

Gradient of a straight line

The gradient measures how steeply the line rises per unit you move right. Between two points (x₁, y₁) and (x₂, y₂) on the line it is the rise over the run. Positive gradient slopes up; negative slopes down; zero is horizontal; undefined (run = 0) is vertical.

The gradient is the rise over the run between any two points on the line.

Straight-line equations (formula booklet) m = y₂ − y₁x₂ − x₁; y = mx + c; y − y₁ = m(x − x₁); ax + by + d = 0

The three equations of a straight line

The three forms describe the same line, just packaged differently. Gradient-intercept y = mx + c reads off the gradient and y-intercept directly. Point-gradient y − y₁ = m(x − x₁) is the natural starting form when you have a point and a gradient. General form ax + by + d = 0 is preferred for integer coefficients and gives both intercepts in one rearrangement.

Match the required form: IB questions almost always specify y = mx + c or ax + by + d = 0. For the general form, multiply through to clear fractions and check the coefficients are integers.

Finding the equation of a straight line

You need a gradient and one point. If given two points, compute the gradient first. Then drop into point-gradient form and rearrange to whichever form the question requests. As a cross-check, the GDC’s linear regression on the two points returns y = ax + b directly.

🧭 Recipe — finding the equation of a straight line

Get the gradient: use m = (y₂ − y₁) / (x₂ − x₁) if two points are given.
Pick a point on the line.
Substitute into y − y₁ = m(x − x₁).
Rearrange into the requested form.
For ax + by + d = 0: multiply through to clear fractions; check integer coefficients.

Worked examples

WE 1

Gradient from two points

Find the gradient of the line through (3, −1) and (7, 5).

apply m = (y₂ − y₁)/(x₂ − x₁) m = (5 − (−1)) / (7 − 3) = 6 / 4 m = 3/2 subtract y-values, then x-values, in the same order.

WE 2

Rearrange to read off gradient and y-intercept

For the line 3x + 2y = 12, find the gradient and the y-intercept.

isolate y 2y = −3x + 12 y = −(3/2)x + 6 compare with y = mx + c m = −3/2 · y-intercept (0, 6) isolate y first — coefficient of x is m, constant is c.

WE 3

Equation from gradient and a point

A line has gradient m = 1/2 and passes through (−4, 3). Find its equation in the form y = mx + c.

point-gradient form y − 3 = (1/2)(x − (−4)) = (1/2)(x + 4) expand & isolate y y − 3 = x/2 + 2 y = x/2 + 5 y = x/2 + 5 check: at x = −4, y = −2 + 5 = 3 ✓

WE 4

Equation from two points (y = mx + c form)

Points A(1, 4) and B(5, −8). Find the equation of line AB in the form y = mx + c.

gradient first m = (−8 − 4)/(5 − 1) = −12/4 = −3 point-gradient with point A y − 4 = −3(x − 1) y − 4 = −3x + 3 y = −3x + 7 check with B: −3(5) + 7 = −8 ✓

WE 5

Equation in ax + by + d = 0 with integer coefficients

Line l passes through (−3, 4) and (5, −2). Find its equation in the form ax + by + d = 0 with a, b, d integers.

gradient m = (−2 − 4)/(5 − (−3)) = −6/8 = −3/4 point-gradient with (−3, 4) y − 4 = (−3/4)(x + 3) multiply both sides by 4 to clear the fraction 4(y − 4) = −3(x + 3) 4y − 16 = −3x − 9 3x + 4y − 7 = 0 check: (−3, 4): −9 + 16 − 7 = 0 ✓; (5, −2): 15 − 8 − 7 = 0 ✓

WE 6

From general form: intercepts and gradient

The line l has equation 4x − 3y + 12 = 0. (a) Find the x– and y-intercepts. (b) Find the gradient. (c) Write l in the form y = mx + c.

(a) x-intercept: set y = 0 4x + 12 = 0 ⇒ x = −3 ⇒ (−3, 0) y-intercept: set x = 0 −3y + 12 = 0 ⇒ y = 4 ⇒ (0, 4) (b) and (c) rearrange to y = mx + c 3y = 4x + 12 ⇒ y = (4/3)x + 4 x-int (−3, 0) · y-int (0, 4) · m = 4/3 · y = (4/3)x + 4 shortcut: x-int = −d/a = −12/4 = −3; y-int = −d/b = 4 ✓

💡 Top tips

Always state in the requested form — y = mx + c or ax + by + d = 0.
For integer coefficients, multiply through by the lowest common denominator before rearranging.
The point-gradient form is your universal starting point; rearrange from there.
Use the GDC’s linear regression on two points as a fast cross-check.
For general form, the shortcuts x-int = −d/a, y-int = −d/b save a rearrangement.

⚠ Common mistakes

Subtracting the y’s and x’s in opposite orders — the result flips sign.
Sign slips on y − (−y₁) and similar — brackets are your friend.
Leaving fractional coefficients in ax + by + d = 0 when integers were asked for.
Forgetting the coefficient of x when reading off the gradient from a rearranged equation.
Mis-reading the y-intercept from ax + by + d = 0 without rearranging first.

Next up: Parallel & Perpendicular Lines — once you have the gradient of one line, the gradient of a parallel or perpendicular line is one quick step away.

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