Write linear equations in two variables in various forms ($y = mx + b$, $Ax + By = C$, $y - y_1 = m(x - x_1)$) given one point and the slope, two points, a table, a graph, or a verbal description (TEKS A.2B, A.2C, A.2G).

What this topic is asking

Writing the equation of a line is central to both linear reporting categories. TEKS A.2B, A.2C, and A.2G ask you to produce a linear equation in any of the three reference-sheet forms ($y = mx + b$, $Ax + By = C$, $y - y_1 = m(x - x_1)$) from whatever information you are given: a point and a slope, two points, a table, a graph, or a verbal description. The skill is choosing the most efficient starting form and converting to whatever the question wants.

Slope and a point: use point-slope form

Point-slope form is the workhorse, because it accepts any point, not just the intercept.

With slope $m = -2$ through $(3, 5)$: $y - 5 = -2(x - 3)$. Distribute and solve for $y$: $y - 5 = -2x + 6$, so $y = -2x + 11$. Point-slope is also the safest choice when the given point is not the $y$-intercept, where dropping the slope straight into $b$ would be wrong.

Two points: slope first, then point-slope

Given two points, compute the slope, then feed either point into point-slope form.

Table, graph, and verbal description

From a table, the slope is the constant change in $y$ over the change in $x$, and the $y$-intercept is the value at $x = 0$ (extend the pattern back if needed). From a graph, read the $y$-intercept and a slope step. From a verbal description, translate: a starting amount is the $y$-intercept, and a per-unit rate is the slope. "A gym charges a $50 joining fee plus$20 a month" becomes $C = 20m + 50$.

Converting to standard form

To reach standard form $Ax + By = C$, clear fractions, move both variables to the left, and make $A$ a non-negative integer. From $y = \frac{2}{3}x - 4$: multiply by 3 to get $3y = 2x - 12$, then $-2x + 3y = -12$, and multiply by $-1$ so $A \ge 0$: $2x - 3y = 12$.

How STAAR examines this topic

• Multiple choice. Pick the correct equation from a point and slope or from a graph; slope-intercept swaps are the standard distractor.
• Equation editor. Build the equation, usually in slope-intercept form, so solve for $y$ before submitting.
• Drag and drop. Drag the slope and intercept values into a form template.

A clarifying idea is that every form describes the same line, so a useful check is to substitute a known point: if the coordinates satisfy your equation, the line is right regardless of which form you used to build it.

Why point-slope is the safest starting point

Slope-intercept form is convenient only when you actually know the $y$-intercept. Point-slope works from any point and slope, which is why it is the reliable default: you never have to first find $b$, and you avoid the error of treating an arbitrary point as the intercept. Once you have $y - y_1 = m(x - x_1)$, a single distribute-and-collect pass converts it to slope-intercept form, and one more rearrangement reaches standard form. Building the habit of "slope, then point-slope, then convert" means a single procedure handles every prompt in this part of the category, whether the information arrives as two points, a graph, or a sentence.

A subtle case is a horizontal or vertical line. A horizontal line through $(3, 5)$ is $y = 5$ (slope 0), and a vertical line through $(3, 5)$ is $x = 3$ (undefined slope, not writable as $y = mx + b$). Point-slope handles the horizontal case directly with $m = 0$, while the vertical case must be written as $x = c$, a distinction the test occasionally probes.

Try this

Q1. Write the line through $(0, 5)$ with slope $-3$. [1 point]

• Cue. $y = -3x + 5$ (the point is the $y$-intercept).

Q2. Write the line through $(1, 4)$ and $(3, 10)$ in slope-intercept form. [2 points]

• Cue. $m = 3$; $y - 4 = 3(x - 1) \Rightarrow y = 3x + 1$.