Linear Functions - SAT Math

$\frac{3}{2}$. Rearrange to get $y = \frac{3}{2}x - 3$.

1. No vertical change means zero slope.

$4x - y = 3$. Move $y$ to left side and arrange coefficients properly.

1. $m = \frac{6-2}{3-1} = \frac{4}{2} = 2$

1. Set $y = 0$ and solve for $x$.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Rise over run between two points $(x_1, y_1)$ and $(x_2, y_2)$.

$2x + y = 5$. Add $2x$ to both sides to get standard form.

The slopes are equal. Lines with identical slopes never intersect.

Y-intercept of the line. The constant term where the line crosses the y-axis.

$y = \frac{3}{2}x - 3$. Solve for $y$ by isolating it on one side.

1. Set $y = 0$ and solve: $0 = 2x - 4$, so $x = 2$.

$Ax + By = C$. General form with integer coefficients $A$, $B$, and $C$.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Change in $y$ divided by change in $x$.

-3. The coefficient of $x$ in slope-intercept form.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Rise over run between two coordinate points.

$y = -\frac{3}{4}x + 3$. Isolate $y$ by dividing both sides by 4.

1. The constant term when $x = 0$.

1. Set $x = 0$ to get $3y = 15$, so $y = 5$.

$x = 5$. Vertical lines have constant $x$-values.

1. Set $y = 0$ to get $2x = 10$, so $x = 5$.

1. Divide by 3: $y = 4x + 2$, so y-intercept is 2.

-3. The coefficient of $x$ is the slope.

Product of slopes is -1. Negative reciprocals create 90-degree intersections.

$y = 3x - 4$. Direct substitution into $y = mx + b$.

$(3, 4)$. Use midpoint formula with given coordinates.

$\frac{1}{2}$. Parallel lines have identical slopes.

Slope of the line. The coefficient of $x$ that determines steepness and direction.

$-\frac{2}{3}$. The coefficient of $x$ in slope-intercept form.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$y - y_1 = m(x - x_1)$. Uses a known point $(x_1, y_1)$ and slope $m$.

1. Use $m = \frac{7-3}{4-2} = \frac{4}{2} = 2$.

$y - y_1 = m(x - x_1)$. Uses a known point $(x_1, y_1)$ and slope $m$.

$y - y_1 = m(x - x_1)$. Uses a known point $(x_1, y_1)$ and slope $m$.

Undefined. Division by zero makes slope undefined.

-7. The constant term in slope-intercept form.

$y = 2x + 8$. Vertical shift changes only the y-intercept.

$Ax + By = C$. General form where $A$, $B$, and $C$ are constants.

Yes, they are parallel. Both lines have slope 2, so they're parallel.

Yes, they are perpendicular. Slopes are 3 and $-\frac{1}{3}$; their product is -1.

$y - 2 = 4(x - 1)$. Substitute point coordinates and slope into formula.

$\bigg( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \bigg)$. Average of coordinates for both $x$ and $y$ values.

$Ax + By = C$. General form with constants $A$, $B$, and $C$.

1. Set $y = 0$ to get $4x = 16$, so $x = 4$.

$y = -(x - 2) + 2$. Horizontal shift affects the $x$-term inside parentheses.

$y + 3 = 2(x - 1)$. Substitute the point and slope into point-slope form.

$y = 7$. Horizontal lines have constant $y$-values.

1. The constant term is the y-intercept.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$y = mx + b$. Standard form where $m$ is slope and $b$ is y-intercept.

$Ax + By = C$. General form with integer coefficients $A$, $B$, and $C$.

$Ax + By = C$. General form where $A$, $B$, and $C$ are constants.

Product of slopes is -1. Negative reciprocals create 90-degree intersections.

$y = 3x - 4$. Direct substitution into $y = mx + b$.

$y = -\frac{3}{4}x + 3$. Isolate $y$ by dividing both sides by 4.

$x = 5$. Vertical lines have constant $x$-values.

$4x - y = 3$. Move $y$ to left side and arrange coefficients properly.

$y = 7$. Horizontal lines have constant $y$-values.

$-\frac{2}{3}$. The coefficient of $x$ in slope-intercept form.

1. Set $y = 0$ to get $2x = 10$, so $x = 5$.

$y = 2x + 8$. Vertical shift changes only the y-intercept.

-3. The coefficient of $x$ is the slope.

1. Set $y = 0$ and solve: $0 = 2x - 4$, so $x = 2$.

1. The constant term is the y-intercept.

$y - y_1 = m(x - x_1)$. Uses a known point $(x_1, y_1)$ and slope $m$.

1. $m = \frac{6-2}{3-1} = \frac{4}{2} = 2$

1. Divide by 3: $y = 4x + 2$, so y-intercept is 2.

1. Use $m = \frac{7-3}{4-2} = \frac{4}{2} = 2$.

$\frac{3}{2}$. Rearrange to get $y = \frac{3}{2}x - 3$.

1. Set $x = 0$ to get $3y = 15$, so $y = 5$.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Change in $y$ divided by change in $x$.

Yes, they are parallel. Both lines have slope 2, so they're parallel.

1. Set $y = 0$ to get $4x = 16$, so $x = 4$.

1. No vertical change means zero slope.

Undefined. Division by zero makes slope undefined.

$2x + y = 5$. Add $2x$ to both sides to get standard form.

$Ax + By = C$. General form with constants $A$, $B$, and $C$.

$m = \frac{y_2 - y_1}{x_2 - x_1}$. Rise over run between two points $(x_1, y_1)$ and $(x_2, y_2)$.

Slope of the line. The coefficient of $x$ that determines steepness and direction.

-3. The coefficient of $x$ in slope-intercept form.

-7. The constant term in slope-intercept form.

Yes, they are perpendicular. Slopes are 3 and $-\frac{1}{3}$; their product is -1.

The slopes are equal. Lines with identical slopes never intersect.

$\bigg( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \bigg)$. Average of coordinates for both $x$ and $y$ values.

$y = -(x - 2) + 2$. Horizontal shift affects the $x$-term inside parentheses.

$(3, 4)$. Use midpoint formula with given coordinates.

$y - y_1 = m(x - x_1)$. Uses a known point $(x_1, y_1)$ and slope $m$.

$y - 2 = 4(x - 1)$. Substitute point coordinates and slope into formula.

Y-intercept of the line. The constant term where the line crosses the y-axis.