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The equation of line CD is [tex]\( y = 3x - 3 \)[/tex].

Write an equation of a line perpendicular to line CD in slope-intercept form that contains point [tex]\( (3, 1) \)[/tex].

A. [tex]\( y = 3x + 0 \)[/tex]
B. [tex]\( y = 3x - 8 \)[/tex]
C. [tex]\( y = \frac{-1}{3}x + 2 \)[/tex]
D. [tex]\( y = \frac{-1}{3}x + 0 \)[/tex]

Sagot :

To find the equation of a line perpendicular to line [tex]\(CD\)[/tex] that passes through a given point [tex]\((3,1)\)[/tex], we need to follow these steps:

### Step 1: Determine the slope of the given line [tex]\(CD\)[/tex]
The equation of line [tex]\(CD\)[/tex] is given by:
[tex]\[ y = 3x - 3 \][/tex]

This is in slope-intercept form, [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope. Therefore, the slope [tex]\( m \)[/tex] of the line [tex]\(CD\)[/tex] is:
[tex]\[ m = 3 \][/tex]

### Step 2: Find the slope of the perpendicular line
The slope of a line that is perpendicular to another line is the negative reciprocal of the slope of the original line. Thus, the slope [tex]\( m' \)[/tex] of the line perpendicular to line [tex]\(CD\)[/tex] is:
[tex]\[ m' = -\frac{1}{3} \][/tex]

### Step 3: Use the point-slope form to find the equation of the perpendicular line
The point-slope form of the equation of a line is:
[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 of the line.

Given that the line passes through the point [tex]\((3, 1)\)[/tex], we substitute [tex]\( (x_1, y_1) = (3, 1) \)[/tex] and [tex]\( m' = -\frac{1}{3} \)[/tex] into the point-slope form:
[tex]\[ y - 1 = -\frac{1}{3}(x - 3) \][/tex]

### Step 4: Simplify the equation to slope-intercept form
First, distribute the slope on the right-hand side:
[tex]\[ y - 1 = -\frac{1}{3}x + 1 \][/tex]

Next, isolate [tex]\( y \)[/tex] to get the equation in slope-intercept form ([tex]\( y = mx + b \)[/tex]):
[tex]\[ y = -\frac{1}{3}x + 1 + 1 \][/tex]
[tex]\[ y = -\frac{1}{3}x + 2 \][/tex]

### Step 5: Write the final equation of the perpendicular line
The equation of the line in slope-intercept form that is perpendicular to line [tex]\(CD\)[/tex] and passes through the point [tex]\((3, 1)\)[/tex] is:
[tex]\[ y = -\frac{1}{3}x + 2 \][/tex]

### Conclusion
From the given options, the correct equation is:
[tex]\[ \boxed{y = -\frac{1}{3}x + 2} \][/tex]