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A coordinate plane of 4 quadrants shows x and y axes ranging from negative 3 to 3 in increments of 1. A line with arrows at each end intercepts two plotted points, A and B. Point A is plotted at left parenthesis 0 comma negative 1 right parenthesis. Point B is plotted at left parenthesis 1 comma 1 right parenthesis. The coordinates are not labeled.

Write the equation of the line in slope-intercept form.

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To find the equation of the line in slope-intercept form (y = mx + b), we need to determine the slope (m) and the y-intercept (b).

  1. Find the slope (m) between points A (0, -1) and B (1, 1): The formula for slope is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Here, (x1, y1) = (0, -1) and (x2, y2) = (1, 1): \[ m = \frac{1 - (-1)}{1 - 0} = \frac{1 + 1}{1} = \frac{2}{1} = 2 \]

  2. Find the y-intercept (b): We can use one of the points to find b. We'll use point A (0, -1): \[ y = mx + b \] Substituting \(x = 0\) and \(y = -1\): \[ -1 = 2(0) + b \implies -1 = b \]

  3. Write the equation in slope-intercept form: Now we have both the slope (m = 2) and the y-intercept (b = -1). The equation of the line is: \[ y = 2x - 1 \]

Thus, the equation of the line in slope-intercept form is: \[ y = 2x - 1 \]