Questions: The graph shows the percentage of a country's adults on weight-loss diets. a. Use the two points whose coordinates are shown to find the point-slope form of the equation of the line that models the percentage of adults on diets, y, x years after 1987. (Simplify your answer. Type your answer in point-slope form. Use integers or fractions for any numbers in the equation.)

The graph shows the percentage of a country's adults on weight-loss diets. a. Use the two points whose coordinates are shown to find the point-slope form of the equation of the line that models the percentage of adults on diets, y, x years after 1987. (Simplify your answer. Type your answer in point-slope form. Use integers or fractions for any numbers in the equation.)
Transcript text: The graph shows the percentage of a country's adults on weight-loss diets. a. Use the two points whose coordinates are shown to find the point-slope form of the equation of the line that models the percentage of adults on diets, $\mathrm{y}, \mathrm{x}$ years after 1987 . $\square$ (Simplify your answer. Type your answer in point-slope form. Use integers or fractions for any numbers in the equation.)
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Solution

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Solution Steps

Step 1: Find the slope

We are given two points on the line: (8, 14) and (18, 12). The slope \(m\) of a line passing through points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\).

In our case, \((x_1, y_1) = (8, 14)\) and \((x_2, y_2) = (18, 12)\). Therefore, \(m = \frac{12 - 14}{18 - 8} = \frac{-2}{10} = -\frac{1}{5}\).

Step 2: Point-slope form

The point-slope form of a linear equation is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line.

Using the point (8, 14) and the slope \(-\frac{1}{5}\), the point-slope form of the equation is \(y - 14 = -\frac{1}{5}(x - 8)\).

Alternatively, using the point (18, 12) and the slope \(-\frac{1}{5}\), the point-slope form of the equation is \(y - 12 = -\frac{1}{5}(x - 18)\).

Final Answer

\(y - 14 = -\frac{1}{5}(x - 8)\) or \(y - 12 = -\frac{1}{5}(x - 18)\). \(\boxed{y - 14 = -\frac{1}{5}(x - 8)}\)

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