Question

Use the image to answer the question.

An illustration of a coordinate plane with four quadrants shows the x-axis ranging from negative 9 to 9 and the y-axis ranging from negative 11 to 11 in increments of 1. A solid upward line with arrows at both ends is plotted on the graph. The line passes through the plotted points left parenthesis 0 comma negative 3 right parenthesis and left parenthesis 1 comma 0 right parenthesis.

What is the equation of the line in the form y=mx+b
?

Answer 1

To find the equation of the line in the form \( y = mx + b \), we first need to determine the slope (\( m \)) of the line using the two points given, which are \( (0, -3) \) and \( (1, 0) \).

The slope \( m \) can be calculated using the formula:

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

Using the points \( (0, -3) \) as \( (x_1, y_1) \) and \( (1, 0) \) as \( (x_2, y_2) \):

\[
m = \frac{0 - (-3)}{1 - 0} = \frac{3}{1} = 3
\]

Now that we have the slope \( m = 3 \), we can use one of the points to find the y-intercept (\( b \)). We'll use the point \( (0, -3) \):

The equation in slope-intercept form is:

\[
y = mx + b
\]

Substituting \( m = 3 \) and using the point \( (0, -3) \):

\[
-3 = 3(0) + b
\]

Solving for \( b \):

\[
-3 = 0 + b \implies b = -3
\]

Now we can write the equation of the line:

\[
y = 3x - 3
\]

So the final answer for the equation of the line is:

\[
\boxed{y = 3x - 3}
\]