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Writing a Linear Equation from Two Points

SAT Math · Algebra · Updated June 2026

Given two points, you can pin down the whole line: find the slope first, then use one point to get the intercept.

Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.

What the SAT tests here

The slope is $\frac{y_2 - y_1}{x_2 - x_1}$. Once you have it, substitute either point into $y = mx + b$ and solve for $b$.

How to solve one, step by step

Example: the line through $(1, 2)$ and $(3, 8)$.

  1. Slope: $\frac{8 - 2}{3 - 1} = 3$.
  2. Use $(1, 2)$: $2 = 3(1) + b \Rightarrow b = -1$.
  3. Equation: $y = 3x - 1$.

The mistakes that cost points

Practice questions

Try these the way you would on test day, then open the solution to check your method.

Easy
In the xy-plane, line $l$ is perpendicular to $y = -\frac{1}{3}x + 1$ and contains the point $(2, -4)$. What is the equation of $l$?
  • A$y = 3x$
  • B$y = 3x - 10$
  • C$y = -3x + 2$
  • D$y = \frac{1}{3}x - \frac{14}{3}$
Show solution
Answer: B, $y = 3x - 10$. The slope of the given line is $-\frac{1}{3}$, so the slope of any perpendicular line is its negative reciprocal: $3$. Substitute the point $(2, -4)$ into $y = 3x + b$: $-4 = 3(2) + b$, so $b = -10$. Equation: $y = 3x - 10$.
Medium
Line $l$ is perpendicular to $y = 9x - 3$ and passes through the point $(18, -8)$. What is the equation of $l$?
  • A$y = \frac{1}{9}x - 10$
  • B$y = -\frac{1}{9}x$
  • C$y = -9x + 154$
  • D$y = -\frac{1}{9}x - 6$
Show solution
Answer: D, $y = -\frac{1}{9}x - 6$. The slope of the given line is $9$, so the slope of any perpendicular line is its negative reciprocal: $-\frac{1}{9}$. Substitute the point $(18, -8)$ into $y = -\frac{1}{9}x + b$: $-8 = -\frac{1}{9}(18) + b$, which gives $-8 = -2 + b$, so $b = -6$. Equation: $y = -\frac{1}{9}x - 6$.
Hard
Line $l$ is perpendicular to $y = -\frac{3}{4}x - 1$ and passes through the point $(3, 2)$. What is the equation of $l$?
  • A$y = \frac{4}{3}x$
  • B$y = -\frac{4}{3}x + 6$
  • C$y = \frac{3}{4}x - \frac{1}{4}$
  • D$y = \frac{4}{3}x - 2$
Show solution
Answer: D, $y = \frac{4}{3}x - 2$. The slope of the given line is $-\frac{3}{4}$, so the slope of any perpendicular line is its negative reciprocal: $\frac{4}{3}$. Substitute the point $(3, 2)$ into $y = \frac{4}{3}x + b$ and solve for $b$. Equation: $y = \frac{4}{3}x - 2$.

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