To determine which point is a solution to the inequality represented by the line in your graph from (0, 1) to (3, -1), we begin by identifying the equation of the line.
The slope of the line can be calculated as follows:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 1}{3 - 0} = \frac{-2}{3} \]
Thus, the equation of the line in point-slope form (using the point (0, 1)) is:
\[ y - 1 = -\frac{2}{3}(x - 0) \implies y = -\frac{2}{3}x + 1 \]
Next, to determine the inequality represented by the graph, we need to observe whether the region above or below the line is shaded:
- If the region above the line is shaded, the inequality is \( y > -\frac{2}{3}x + 1 \).
- If the region below the line is shaded, the inequality is \( y < -\frac{2}{3}x + 1 \).
Now you will want to check a test point that is either inside the shaded region or outside it to determine if it satisfies the inequality.
For example, let’s test the point (0, 0):
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Substitute \( x = 0 \) into the inequality:
If the inequality is \( y > -\frac{2}{3}x + 1 \): \[ 0 > -\frac{2}{3}(0) + 1 \implies 0 > 1 \quad \text{(False)} \]
If the inequality is \( y < -\frac{2}{3}x + 1 \): \[ 0 < -\frac{2}{3}(0) + 1 \implies 0 < 1 \quad \text{(True)} \]
If the test point is (1, 0):
For \( y > -\frac{2}{3}x + 1 \): \[ 0 > -\frac{2}{3}(1) + 1 \implies 0 > \frac{1}{3} \quad \text{(False)} \]
For \( y < -\frac{2}{3}x + 1 \): \[ 0 < -\frac{2}{3}(1) + 1 \implies 0 < \frac{1}{3} \quad \text{(True)} \]
From this, we can conclude that any point below the line (in the correct region) satisfies the corresponding inequality.
You will need to select the options presented in your question (if any) to find which points lie on the valid area below or above the line based on its representation in your graph.