Question

8.
A line passes through (3, –2) and (6, 2). Write an equation for the line in point-slope form.
Rewrite the equation in standard form using integers.

Answer 1

To write the equation of a line that passes through the points \((3, -2)\) and \((6, 2)\), we can start by calculating the slope (m) of the line using the formula:

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

Substituting the coordinates of the points \((x_1, y_1) = (3, -2)\) and \((x_2, y_2) = (6, 2)\):

\[
m = \frac{2 - (-2)}{6 - 3} = \frac{2 + 2}{6 - 3} = \frac{4}{3}
\]

Now that we have the slope \(m = \frac{4}{3}\), 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 \((3, -2)\) as \((x_1, y_1)\):

\[
y - (-2) = \frac{4}{3}(x - 3)
\]

Simplifying this gives:

\[
y + 2 = \frac{4}{3}(x - 3)
\]

Now we can rewrite it in standard form. First, we can eliminate the fraction by multiplying both sides by 3:

\[
3(y + 2) = 4(x - 3)
\]

Expanding both sides:

\[
3y + 6 = 4x - 12
\]

Now, we can rearrange it into standard form \(Ax + By = C\):

\[
4x - 3y = 18
\]

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

\[
4x - 3y = 18
\]