Find slope and intercept, write the equation of a line from two points or a graph, and interpret what the rate of change actually means in context. Every question comes with a written walkthrough of exactly how the equation was built.
Swapping which point is (x₁,y₁) and which is (x₂,y₂) in only the numerator or only the denominator flips the sign of the slope — the order has to match in both.
A slope of 2/3 is parallel to another line with slope 2/3, but perpendicular to one with slope −3/2 — flipping and negating, not just flipping or just negating.
Plugging a known point into y=mx+b to find the y-intercept requires solving the resulting equation for b — stopping after substitution leaves an equation, not an answer.
Straight from the bank — one per difficulty tier. Reveal the answer to see the explanation you'd get in a real session.
Find the slope of the line through (2, 3) and (6, 11).
A — 2. Slope is the change in y over the change in x: (11−3)/(6−2) = 8/4 = 2. 0.5 (B) comes from inverting the fraction — dividing the change in x by the change in y instead.
A line has a slope of −3 and passes through (4, 2). What is its y-intercept?
A — 14. Substitute the point into y = mx + b: 2 = −3(4) + b, so 2 = −12 + b, giving b = 14. -10 (B) comes from stopping after substitution and combining in the wrong direction — computing 2 − 12 instead of isolating b by adding 12 to both sides.
Line ℓ passes through (1, 5) and is perpendicular to a line with equation y = (2/3)x − 4. What is the equation of line ℓ?
A — y = -3/2 x + 13/2. A perpendicular slope is the negative reciprocal of 2/3, which is −3/2. Substituting the point (1, 5) into y = −3/2 x + b: 5 = −3/2(1) + b, so b = 5 + 3/2 = 13/2, giving y = −3/2 x + 13/2. y = -3/2 x + 7 (C) uses the correct slope but rounds the y-intercept instead of keeping it as the fraction the arithmetic actually produces.
Linear function problems reward knowing which formula to reach for — slope from two points, point-slope, or parallel/perpendicular rules — as much as the arithmetic itself.
Before writing anything, name what you have — two points, a slope and a point, or a parallel/perpendicular condition — and pick the matching approach. Untimed practice is where that judgment gets fast.
Move to timed sessions once formula selection is automatic. Perpendicular and parallel problems run a little longer since there's an extra step before the main equation.
Pair linear functions with quadratic functions in a mock — reading a graph and connecting it to an equation is the shared skill.
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