To find the equation of a line perpendicular to the given line [tex]\( CD \)[/tex] and passing through the point [tex]\((3,1)\)[/tex], follow these steps:
1. Determine the slope of line [tex]\( CD \)[/tex]:
The given equation of line [tex]\( CD \)[/tex] is [tex]\( y = 3x - 3 \)[/tex]. The slope [tex]\( m \)[/tex] of this line is the coefficient of [tex]\( x \)[/tex], which is [tex]\( 3 \)[/tex].
2. Find the slope of the perpendicular line:
Lines that are perpendicular to each other have slopes that are negative reciprocals of each other. The negative reciprocal of [tex]\( 3 \)[/tex] is [tex]\(-\frac{1}{3} \)[/tex]. So, the slope of the perpendicular line is [tex]\(-\frac{1}{3} \)[/tex].
3. Use the point-slope form to write the equation:
The point-slope form of a line's equation 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. We need to use the given point [tex]\((3, 1)\)[/tex] and the slope [tex]\(-\frac{1}{3} \)[/tex] to find the equation.
Plugging in the point and the slope:
[tex]\[ y - 1 = -\frac{1}{3} (x - 3) \][/tex]
4. Simplify to slope-intercept form [tex]\( y = mx + b \)[/tex]:
Let's solve for [tex]\( y \)[/tex] to put the equation in the form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y - 1 = -\frac{1}{3}x + \frac{1}{3}(3) \][/tex]
[tex]\[ y - 1 = -\frac{1}{3}x + 1 \][/tex]
[tex]\[ y = -\frac{1}{3}x + 1 + 1 \][/tex]
[tex]\[ y = -\frac{1}{3}x + 2 \][/tex]
5. Identify the correct choice:
The equation of the perpendicular line that passes through the point [tex]\((3,1)\)[/tex] is [tex]\( y = -\frac{1}{3} x + 2 \)[/tex].
The correct multiple-choice answer is:
[tex]\[ \boxed{y = \frac{-1}{3} x + 2} \][/tex]
