Question

A coordinate plane with a line passing through (0, negative 5) and (3, 0).
What is the equation of the graphed line written in standard form?

3x + y = –5
x + 3y = –5
3x – 5y = –15
5x – 3y = 15

Answer 1

To find the equation of the line passing through the points (0, -5) and (3, 0), we can start by calculating the slope (m) of the line. The slope is given by the formula:

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

Using the points (0, -5) and (3, 0):

- \( x_1 = 0, y_1 = -5 \)
- \( x_2 = 3, y_2 = 0 \)

The slope becomes:

\[
m = \frac{0 - (-5)}{3 - 0} = \frac{5}{3}
\]

Now that we have the slope, we can use the point-slope form of the equation of a line, which is:

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

Using the point (0, -5):

\[
y - (-5) = \frac{5}{3}(x - 0)
\]

Simplifying this gives:

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

Subtracting 5 from both sides:

\[
y = \frac{5}{3}x - 5
\]

To convert this equation to standard form \(Ax + By = C\), we can multiply everything by 3 to eliminate the fraction:

\[
3y = 5x - 15
\]

Rearranging it gives:

\[
-5x + 3y = -15
\]

To put it in the standard form \(Ax + By = C\) with A positive, multiply the entire equation by -1:

\[
5x - 3y = 15
\]

Thus, the equation of the line in standard form is:

\[
5x - 3y = 15
\]

So the correct answer is:

**5x - 3y = 15**