To find the equation of the line that passes through the points [tex]\((-3, -1)\)[/tex] and [tex]\((3, 3)\)[/tex], we can follow these steps:
1. Calculate the Slope of the Line ([tex]\(m\)[/tex]):
The formula for the slope [tex]\(m\)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Using the points [tex]\((-3, -1)\)[/tex] and [tex]\((3, 3)\)[/tex], we get:
[tex]\[
m = \frac{3 - (-1)}{3 - (-3)} = \frac{3 + 1}{3 + 3} = \frac{4}{6} = \frac{2}{3}
\][/tex]
2. Form the Equation in Point-Slope Form:
The point-slope form of the equation of a line is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Substituting [tex]\(m = \frac{2}{3}\)[/tex], [tex]\(x_1 = -3\)[/tex], and [tex]\(y_1 = -1\)[/tex]:
[tex]\[
y - (-1) = \frac{2}{3}(x - (-3))
\][/tex]
Simplify:
[tex]\[
y + 1 = \frac{2}{3}(x + 3)
\][/tex]
3. Convert to Slope-Intercept Form ([tex]\(y = mx + b\)[/tex]):
Distribute [tex]\(\frac{2}{3}\)[/tex] and simplify:
[tex]\[
y + 1 = \frac{2}{3}x + \frac{2}{3} \cdot 3
\][/tex]
[tex]\[
y + 1 = \frac{2}{3}x + 2
\][/tex]
Subtract 1 from both sides:
[tex]\[
y = \frac{2}{3}x + 1
\][/tex]
4. Convert to Standard Form [tex]\(Ax + By = C\)[/tex]:
Multiply all terms by 3 to eliminate the fraction:
[tex]\[
3y = 2x + 3
\][/tex]
Rearrange to get the standard form:
[tex]\[
2x - 3y = -3
\][/tex]
So, the equation of the line that goes through [tex]\((-3, -1)\)[/tex] and [tex]\((3, 3)\)[/tex] is:
[tex]\[2x - 3y = -3\][/tex]
Therefore, the correct option is:
D. [tex]\(2x - 3y = -3\)[/tex]
