To find the equation of a straight line that passes through the point [tex]\((3, 5)\)[/tex] and is perpendicular to the given line [tex]\(y = 3x + 2\)[/tex], follow these steps:
### Step 1: Determine the Slope of the Given Line
The equation of the given line is in the slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope. Here, the slope [tex]\(m\)[/tex] of the given line [tex]\(y = 3x + 2\)[/tex] is [tex]\(3\)[/tex].
### Step 2: Find the Slope of the Perpendicular Line
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. Thus, if the slope of the given line is [tex]\(3\)[/tex], the slope [tex]\(m'\)[/tex] of the line perpendicular to it is:
[tex]\[
m' = -\frac{1}{3}
\][/tex]
### Step 3: Use the Point-Slope Form to Write the Equation
We know that the line passes through the point [tex]\((3, 5)\)[/tex] and has a slope of [tex]\(-\frac{1}{3}\)[/tex]. We use the point-slope form of the line equation, which is given by:
[tex]\[
y - y_1 = m'(x - x_1)
\][/tex]
Substitute [tex]\(x_1 = 3\)[/tex], [tex]\(y_1 = 5\)[/tex], and [tex]\(m' = -\frac{1}{3}\)[/tex] into the equation:
[tex]\[
y - 5 = -\frac{1}{3}(x - 3)
\][/tex]
### Step 4: Simplify the Equation
Next, simplify the equation to get it in slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[
y - 5 = -\frac{1}{3}(x - 3)
\][/tex]
Distribute the [tex]\(-\frac{1}{3}\)[/tex] on the right side:
[tex]\[
y - 5 = -\frac{1}{3}x + \frac{1}{3} \cdot 3
\][/tex]
[tex]\[
y - 5 = -\frac{1}{3}x + 1
\][/tex]
Add 5 to both sides to isolate [tex]\(y\)[/tex]:
[tex]\[
y = -\frac{1}{3}x + 1 + 5
\][/tex]
[tex]\[
y = -\frac{1}{3}x + 6
\][/tex]
### Final Equation
The equation of the line that passes through the point [tex]\((3, 5)\)[/tex] and is perpendicular to the line [tex]\(y = 3x + 2\)[/tex] is:
[tex]\[
y = -\frac{1}{3}x + 6
\][/tex]
So, the line has a slope of [tex]\(-\frac{1}{3}\)[/tex] and its equation is [tex]\(y = -\frac{1}{3}x + 6\)[/tex].
