We are asked to find the equation of the line shown in the graph.

AlgebraLinear EquationsSlopePoint-Slope FormCoordinate Geometry
2025/7/3

1. Problem Description

We are asked to find the equation of the line shown in the graph.

2. Solution Steps

First, we need to find two points on the line. From the graph, we can identify the following two points:
Point 1: (2,2)(-2, -2)
Point 2: (8,4)(8, -4)
Next, we will calculate the slope mm of the line using the formula:
m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}
Plugging in the coordinates of the points, we get:
m=4(2)8(2)=4+28+2=210=15m = \frac{-4 - (-2)}{8 - (-2)} = \frac{-4 + 2}{8 + 2} = \frac{-2}{10} = -\frac{1}{5}
Now that we have the slope, we can use the point-slope form of a linear equation:
yy1=m(xx1)y - y_1 = m(x - x_1)
Using Point 1 (2,2)(-2, -2) and the slope m=15m = -\frac{1}{5}, we get:
y(2)=15(x(2))y - (-2) = -\frac{1}{5}(x - (-2))
y+2=15(x+2)y + 2 = -\frac{1}{5}(x + 2)
y+2=15x25y + 2 = -\frac{1}{5}x - \frac{2}{5}
y=15x252y = -\frac{1}{5}x - \frac{2}{5} - 2
y=15x25105y = -\frac{1}{5}x - \frac{2}{5} - \frac{10}{5}
y=15x125y = -\frac{1}{5}x - \frac{12}{5}

3. Final Answer

The equation of the line is y=15x125y = -\frac{1}{5}x - \frac{12}{5}.

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