To determine the equation of the line that passes through the origin and is perpendicular to the line passing through the points [tex]\(A(-3, 0)\)[/tex] and [tex]\(B(-6, 5)\)[/tex], follow these steps:
1. Find the slope of the line passing through points [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
The formula for the slope [tex]\(m\)[/tex] of a line between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
For points [tex]\(A(-3, 0)\)[/tex] and [tex]\(B(-6, 5)\)[/tex]:
[tex]\[
m_{AB} = \frac{5 - 0}{-6 - (-3)} = \frac{5}{-3} = -\frac{5}{3}
\][/tex]
2. Determine the slope of the line perpendicular to the line passing through [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
If two lines are perpendicular, the product of their slopes is [tex]\(-1\)[/tex]:
[tex]\[
m_{AB} \times m_{\perp} = -1
\][/tex]
Given the slope of the line passing through [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is [tex]\(-\frac{5}{3}\)[/tex], the slope of the perpendicular line [tex]\(m_{\perp}\)[/tex] is:
[tex]\[
m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{5}{3}} = \frac{3}{5}
\][/tex]
3. Form the equation of the line passing through the origin with the perpendicular slope:
Since the line passes through the origin [tex]\((0, 0)\)[/tex] and has a slope [tex]\(\frac{3}{5}\)[/tex], we can use the slope-intercept form of the equation of a line:
[tex]\[
y = mx + c
\][/tex]
Here, [tex]\(m = \frac{3}{5}\)[/tex] and [tex]\(c = 0\)[/tex] (since it passes through the origin):
[tex]\[
y = \frac{3}{5}x
\][/tex]
To express this equation in standard form [tex]\(Ax + By = C\)[/tex], multiply both sides by 5:
[tex]\[
5y = 3x
\][/tex]
Rearrange to get standard form:
[tex]\[
3x - 5y = 0
\][/tex]
Thus, the equation of the line that passes through the origin and is perpendicular to the line passing through points [tex]\(A(-3, 0)\)[/tex] and [tex]\(B(-6, 5)\)[/tex] is:
[tex]\[
3x - 5y = 0
\][/tex]
