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What is the equation, in slope-intercept form, of the line that is perpendicular to the given line and passes through the point (2, -1)?

A. [tex]\( y = -\frac{1}{3} x - \frac{1}{3} \)[/tex]

B. [tex]\( y = -\frac{1}{3} x - \frac{5}{3} \)[/tex]

C. [tex]\( y = 3x - 3 \)[/tex]

D. [tex]\( y = 3x - 7 \)[/tex]

Sagot :

To find the equation of a line that is perpendicular to a given line and passes through a specific point, follow these steps:

1. Identify the slope of the given line:
The given equation of the line is [tex]\( y = -\frac{1}{3}x - \frac{5}{3} \)[/tex]. The slope ([tex]\(m\)[/tex]) of this line is [tex]\(-\frac{1}{3}\)[/tex].

2. Determine the slope of the perpendicular line:
The slope of a line that is perpendicular to another line is the negative reciprocal of the original line's slope. The negative reciprocal of [tex]\(-\frac{1}{3}\)[/tex] is [tex]\(3\)[/tex]. So, the slope of the perpendicular line is [tex]\(3\)[/tex].

3. Use the point-slope form of the equation:
The point-slope form of a line equation is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\(m\)[/tex] is the slope of the line, and [tex]\((x_1, y_1)\)[/tex] is a point through which the line passes. We are given the point [tex]\((2, -1)\)[/tex].

4. Plug in the given point and the slope:
Substitute [tex]\(m = 3\)[/tex], [tex]\(x_1 = 2\)[/tex], and [tex]\(y_1 = -1\)[/tex] into the point-slope form equation:
[tex]\[ y - (-1) = 3(x - 2) \][/tex]
This simplifies to:
[tex]\[ y + 1 = 3(x - 2) \][/tex]

5. Simplify the equation to slope-intercept form:
Distribute the [tex]\(3\)[/tex] on the right-hand side:
[tex]\[ y + 1 = 3x - 6 \][/tex]
Subtract [tex]\(1\)[/tex] from both sides to solve for [tex]\(y\)[/tex]:
[tex]\[ y = 3x - 7 \][/tex]

Thus, the equation of the line that is perpendicular to [tex]\(y = -\frac{1}{3}x - \frac{5}{3}\)[/tex] and passes through the point [tex]\((2, -1)\)[/tex] is:
[tex]\[ y = 3x - 7 \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{y = 3x - 7} \][/tex]
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