Sure! To find the equation of the straight line passing through the point [tex]\((3,5)\)[/tex] which is perpendicular to the line [tex]\(y = 3x + 2\)[/tex], we need to follow these steps:
1. Identify the slope of the given line:
The equation of the given line is [tex]\(y = 3x + 2\)[/tex]. This is in the slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope. From this equation, we can see that the slope [tex]\(m\)[/tex] of the given line is [tex]\(3\)[/tex].
2. Determine the slope of the perpendicular line:
The slope of a line perpendicular to another is the negative reciprocal of the slope of the original line. The negative reciprocal of [tex]\(3\)[/tex] is [tex]\(-\frac{1}{3}\)[/tex].
3. Use the point-slope form of the equation of a line:
The point-slope form of the equation of a line is given by:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line and [tex]\(m\)[/tex] is the slope.
4. Substitute the given point and the slope of the perpendicular line into the point-slope form:
The given point is [tex]\((3, 5)\)[/tex] and the slope of the perpendicular line is [tex]\(-\frac{1}{3}\)[/tex]. Substituting these into the point-slope form, we get:
[tex]\[
y - 5 = -\frac{1}{3}(x - 3)
\][/tex]
5. Simplify to find the equation in slope-intercept form:
To simplify, distribute the slope on the right-hand side:
[tex]\[
y - 5 = -\frac{1}{3}x + 1
\][/tex]
Next, isolate [tex]\(y\)[/tex] by adding [tex]\(5\)[/tex] to both sides:
[tex]\[
y = -\frac{1}{3}x + 1 + 5
\][/tex]
[tex]\[
y = -\frac{1}{3}x + 6
\][/tex]
6. Conclusion:
The equation of the line passing through the point [tex]\((3, 5)\)[/tex] and perpendicular to the line [tex]\(y = 3x + 2\)[/tex] is:
[tex]\[
y = -\frac{1}{3}x + 6
\][/tex]
So, the solution leads us to the equation [tex]\(y = -\frac{1}{3}x + 6\)[/tex], which is the required equation of the line.
