Write the equation of a line in slope-intercept and point-slope form given a slope and a point, two points, or a graph (LA A1: F-IF, A-CED, building linear models).

What this topic is asking

This dot point asks you to write the equation of a line from a slope and a point, from two points, or from a graph, in slope-intercept or point-slope form. On LEAP 2025 these are Type I Major Content items, central to linear modeling. As with slope, the line forms are not on the reference sheet, so know them.

From a slope and a point: point-slope form

When you know the slope and one point, point-slope form is the most direct.

From two points

With two points, compute the slope first, then use point-slope with either point (the result is the same).

Using the other point $(3, 9)$ gives $y - 9 = 2(x - 3) = 2x - 6$, so $y = 2x + 3$, the same line.

Reading the y-intercept directly

If one of your points has $x = 0$, that point is the $y$-intercept: its $y$-value is $b$. Then you only need the slope to write $y = mx + b$ in one step, no point-slope needed.

How LEAP examines this topic

• Equation response. Write the equation in a requested form and enter it.
• Multiple choice. Pick the correct equation from a slope and point or two points.
• Graphing item. Read slope and a point off a graph, then write the equation (or the reverse).

A clarifying idea: point-slope and slope-intercept describe the same line; you can always convert point-slope to slope-intercept by distributing and solving for $y$. Choose whichever the item asks for.

Why two forms exist for one line

Having both point-slope and slope-intercept form is a matter of convenience matched to the information you start with, which is the practical point of this standard. Slope-intercept form $y = mx + b$ requires the $y$-intercept, the value at $x = 0$, which you may not be given. Point-slope form $y - y_1 = m(x - x_1)$ requires only any point and the slope, so it works directly whenever you have a slope and a coordinate, with no need to first find where the line crosses the $y$-axis. The two forms are algebraically identical: distributing point-slope and solving for $y$ always lands on slope-intercept form. The reason to learn both is speed and reliability, starting from point-slope when you have an arbitrary point avoids the extra step of solving for $b$, and starting from slope-intercept is fastest when the intercept is handed to you. Recognizing which form the given information fits is the judgment the test rewards, and it is why a two-point problem always begins by computing the slope: the slope is the one ingredient both forms require.

Try this

Q1. Write the line with slope $3$ through $(2, 1)$ in slope-intercept form. [2 points]

• Cue. $y - 1 = 3(x - 2) \Rightarrow y = 3x - 5$.

Q2. Write the line through $(0, 4)$ and $(2, 0)$. [2 points]

• Cue. $m = \frac{0 - 4}{2 - 0} = -2$; intercept $4$, so $y = -2x + 4$.