Question

What is the equation of the line that passes through the points (−3,4) and (−6,3)? Write the answer in slope-intercept form.
y=(?)x+?

Answer 1

To find the equation of the line that passes through the points \((-3, 4)\) and \((-6, 3)\), we first need to calculate the slope of the line using the formula:

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

Let \((x_1, y_1) = (-3, 4)\) and \((x_2, y_2) = (-6, 3)\). Plugging in these values:

\[
m = \frac{3 - 4}{-6 - (-3)} = \frac{-1}{-6 + 3} = \frac{-1}{-3} = \frac{1}{3}
\]

Now we have the slope \(m = \frac{1}{3}\).

Next, we can use the point-slope form of the equation of a line, which is given by:

\[
y - y_1 = m(x - x_1)
\]

Using one of the points, say \((-3, 4)\), we can substitute in the values:

\[
y - 4 = \frac{1}{3}(x + 3)
\]

Now we will simplify this equation:

\[
y - 4 = \frac{1}{3}x + 1
\]

Adding 4 to both sides:

\[
y = \frac{1}{3}x + 1 + 4
\]
\[
y = \frac{1}{3}x + 5
\]

Thus, in slope-intercept form \(y = mx + b\), the equation of the line is:

\[
y = \frac{1}{3}x + 5
\]

So, your answer is:

\[
y = \frac{1}{3}x + 5
\]