Question

The trend line starts at the point 15, 40 and ends at 30, 60
Write an equation in slope-intercept form of the trend line.

Answer 1

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

1. **Calculate the slope \( m \)**:

The formula for the slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

Given the points \((15, 40)\) and \((30, 60)\):

\[
m = \frac{60 - 40}{30 - 15} = \frac{20}{15} = \frac{4}{3}
\]

2. **Use the slope and one point to find the y-intercept \( b \)**:

We can use the point-slope form of the equation and rewrite it in slope-intercept form. We have \( m = \frac{4}{3} \) and we can use the point \((15, 40)\):

\[
y - y_1 = m(x - x_1)
\]
\[
y - 40 = \frac{4}{3}(x - 15)
\]

Expanding this equation:

\[
y - 40 = \frac{4}{3}x - \frac{4}{3} \cdot 15
\]
\[
y - 40 = \frac{4}{3}x - 20
\]
\[
y = \frac{4}{3}x - 20 + 40
\]
\[
y = \frac{4}{3}x + 20
\]

Thus, the equation of the trend line in slope-intercept form is:

\[
y = \frac{4}{3}x + 20
\]