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To find the equation of the line that is perpendicular to the given lines and passes through the point [tex]\((2, -1)\)[/tex], we need to follow a series of steps. Let's break it down step by step:
1. Determine the slope of the given lines:
The given lines have the equations [tex]\(y = -\frac{1}{3}x - \frac{1}{3}\)[/tex] and [tex]\(y = -\frac{1}{3}x - \frac{5}{3}\)[/tex].
2. Identify the slope of the perpendicular line:
A line perpendicular to another line has a slope that is the negative reciprocal of the original line's slope. The slope of the given lines is [tex]\(-\frac{1}{3}\)[/tex]. Therefore, the slope of the perpendicular line will be the negative reciprocal of [tex]\(-\frac{1}{3}\)[/tex]:
[tex]\[ \text{slope of the perpendicular line} = -\left(\frac{1}{-\frac{1}{3}}\right) = 3. \][/tex]
3. Use the point-slope form to find the equation of the line:
The point-slope form of a line equation is given by:
[tex]\[ y - y_1 = m(x - x_1), \][/tex]
where [tex]\(m\)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is a point on the line.
Here, the slope [tex]\(m\)[/tex] is [tex]\(3\)[/tex] and the point [tex]\((x_1, y_1)\)[/tex] is [tex]\((2, -1)\)[/tex]. Plugging these values into the point-slope form, we get:
[tex]\[ y - (-1) = 3(x - 2). \][/tex]
4. Simplify:
Simplify the equation to convert it to slope-intercept form, [tex]\(y = mx + b\)[/tex]:
[tex]\[ y + 1 = 3(x - 2) \][/tex]
[tex]\[ y + 1 = 3x - 6. \][/tex]
Subtract 1 from both sides to solve for [tex]\(y\)[/tex]:
[tex]\[ y = 3x - 6 - 1 \][/tex]
[tex]\[ y = 3x - 7. \][/tex]
Therefore, the equation of the line that is perpendicular to the given lines and passes through the point [tex]\((2, -1)\)[/tex] is [tex]\(\boxed{y = 3x - 7}\)[/tex].
1. Determine the slope of the given lines:
The given lines have the equations [tex]\(y = -\frac{1}{3}x - \frac{1}{3}\)[/tex] and [tex]\(y = -\frac{1}{3}x - \frac{5}{3}\)[/tex].
2. Identify the slope of the perpendicular line:
A line perpendicular to another line has a slope that is the negative reciprocal of the original line's slope. The slope of the given lines is [tex]\(-\frac{1}{3}\)[/tex]. Therefore, the slope of the perpendicular line will be the negative reciprocal of [tex]\(-\frac{1}{3}\)[/tex]:
[tex]\[ \text{slope of the perpendicular line} = -\left(\frac{1}{-\frac{1}{3}}\right) = 3. \][/tex]
3. Use the point-slope form to find the equation of the line:
The point-slope form of a line equation is given by:
[tex]\[ y - y_1 = m(x - x_1), \][/tex]
where [tex]\(m\)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is a point on the line.
Here, the slope [tex]\(m\)[/tex] is [tex]\(3\)[/tex] and the point [tex]\((x_1, y_1)\)[/tex] is [tex]\((2, -1)\)[/tex]. Plugging these values into the point-slope form, we get:
[tex]\[ y - (-1) = 3(x - 2). \][/tex]
4. Simplify:
Simplify the equation to convert it to slope-intercept form, [tex]\(y = mx + b\)[/tex]:
[tex]\[ y + 1 = 3(x - 2) \][/tex]
[tex]\[ y + 1 = 3x - 6. \][/tex]
Subtract 1 from both sides to solve for [tex]\(y\)[/tex]:
[tex]\[ y = 3x - 6 - 1 \][/tex]
[tex]\[ y = 3x - 7. \][/tex]
Therefore, the equation of the line that is perpendicular to the given lines and passes through the point [tex]\((2, -1)\)[/tex] is [tex]\(\boxed{y = 3x - 7}\)[/tex].
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