Question

What is the equation of this line in standard form?

Responses

8x−9y=23
8 x minus9 y equals 23

9x−8y=23
9 x minus 8 y equals 23

8x−7y=−25
8 x minus 7 y equals negative 25

8x−9y=−23
8 x minus 9 y equals negative 23
Number graph ranging from negative five to five on the x and y axes. A line is drawn on the graph that passes through the labeled points begin ordered pair negative 4 comma negative 1 end ordered pair and begin ordered pair one-half comma 3 end ordered pair

Answer 1

To find the equation of the line in standard form that passes through the points (-4, -1) and (1/2, 3), we first need to determine the slope (m) of the line using the formula:

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

Let \((x_1, y_1) = (-4, -1)\) and \((x_2, y_2) = \left(\frac{1}{2}, 3\right)\):

\[
m = \frac{3 - (-1)}{\frac{1}{2} - (-4)} = \frac{3 + 1}{\frac{1}{2} + 4} = \frac{4}{\frac{1}{2} + \frac{8}{2}} = \frac{4}{\frac{9}{2}} = 4 \cdot \frac{2}{9} = \frac{8}{9}
\]

Now that we have the slope \(m = \frac{8}{9}\), we can use the point-slope form of the equation of a line:

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

Using point \((-4, -1)\):

\[
y - (-1) = \frac{8}{9}(x - (-4))
\]

This simplifies to:

\[
y + 1 = \frac{8}{9}(x + 4)
\]
\[
y + 1 = \frac{8}{9}x + \frac{32}{9}
\]

Now, subtract 1 (which is \(\frac{9}{9}\)) from both sides:

\[
y = \frac{8}{9}x + \frac{32}{9} - \frac{9}{9}
\]
\[
y = \frac{8}{9}x + \frac{23}{9}
\]

Next, we need to convert this equation into standard form \(Ax + By = C\). We first eliminate the fraction by multiplying the entire equation by 9:

\[
9y = 8x + 23
\]

Rearranging gives:

\[
-8x + 9y = 23
\]

We can multiply the entire equation by -1 to put it in standard form:

\[
8x - 9y = -23
\]

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

\[
\boxed{8x - 9y = -23}
\]