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Sagot :
To find the equation of the line passing through the point [tex]\((12, 3)\)[/tex] that is perpendicular to the line [tex]\(y = -3x\)[/tex], we need to follow these steps:
1. Determine the slope of the given line:
The equation of the given line is [tex]\(y = -3x\)[/tex]. Here, the slope ([tex]\(m\)[/tex]) of the line is [tex]\(-3\)[/tex].
2. Find the slope of the perpendicular line:
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the given line. So, the slope of the line perpendicular to [tex]\(y = -3x\)[/tex] is:
[tex]\[ \text{Perpendicular slope} = -\frac{1}{\text{slope of the given line}} = -\frac{1}{-3} = \frac{1}{3} \][/tex]
3. Use the point-slope form of the equation:
The point-slope form of the equation of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\(m\)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the point the line passes through. Substituting [tex]\((x_1, y_1) = (12, 3)\)[/tex] and the slope [tex]\(m = \frac{1}{3}\)[/tex], we get:
[tex]\[ y - 3 = \frac{1}{3}(x - 12) \][/tex]
4. Solve for [tex]\(y\)[/tex] to get the slope-intercept form:
Simplify the equation to put it into the form [tex]\(y = mx + b\)[/tex]:
[tex]\[ y - 3 = \frac{1}{3}x - \frac{1}{3}(12) \][/tex]
[tex]\[ y - 3 = \frac{1}{3}x - 4 \][/tex]
Add 3 to both sides:
[tex]\[ y = \frac{1}{3}x - 4 + 3 \][/tex]
[tex]\[ y = \frac{1}{3}x - 1 \][/tex]
So, the equation of the line passing through the point [tex]\((12, 3)\)[/tex] and perpendicular to [tex]\(y = -3x\)[/tex] is:
[tex]\[ \boxed{y = \frac{1}{3}x - 1} \][/tex]
1. Determine the slope of the given line:
The equation of the given line is [tex]\(y = -3x\)[/tex]. Here, the slope ([tex]\(m\)[/tex]) of the line is [tex]\(-3\)[/tex].
2. Find the slope of the perpendicular line:
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the given line. So, the slope of the line perpendicular to [tex]\(y = -3x\)[/tex] is:
[tex]\[ \text{Perpendicular slope} = -\frac{1}{\text{slope of the given line}} = -\frac{1}{-3} = \frac{1}{3} \][/tex]
3. Use the point-slope form of the equation:
The point-slope form of the equation of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\(m\)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the point the line passes through. Substituting [tex]\((x_1, y_1) = (12, 3)\)[/tex] and the slope [tex]\(m = \frac{1}{3}\)[/tex], we get:
[tex]\[ y - 3 = \frac{1}{3}(x - 12) \][/tex]
4. Solve for [tex]\(y\)[/tex] to get the slope-intercept form:
Simplify the equation to put it into the form [tex]\(y = mx + b\)[/tex]:
[tex]\[ y - 3 = \frac{1}{3}x - \frac{1}{3}(12) \][/tex]
[tex]\[ y - 3 = \frac{1}{3}x - 4 \][/tex]
Add 3 to both sides:
[tex]\[ y = \frac{1}{3}x - 4 + 3 \][/tex]
[tex]\[ y = \frac{1}{3}x - 1 \][/tex]
So, the equation of the line passing through the point [tex]\((12, 3)\)[/tex] and perpendicular to [tex]\(y = -3x\)[/tex] is:
[tex]\[ \boxed{y = \frac{1}{3}x - 1} \][/tex]
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