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To determine the equation of the line that is perpendicular to a given line and passes through a specific point, let's work through the problem step-by-step.
We begin by identifying the slopes of the given lines and then finding the slope of the line that will be perpendicular to them.
1. Identifying the slopes of the given lines:
- The equation [tex]\(3x + 5y = -9\)[/tex] can be rewritten in the slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ 5y = -3x - 9 \][/tex]
[tex]\[ y = -\frac{3}{5}x - \frac{9}{5} \][/tex]
The slope [tex]\(m\)[/tex] is [tex]\(-\frac{3}{5}\)[/tex].
- The equation [tex]\(3x + 5y = 9\)[/tex] can be rewritten in the slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ 5y = -3x + 9 \][/tex]
[tex]\[ y = -\frac{3}{5}x + \frac{9}{5} \][/tex]
The slope [tex]\(m\)[/tex] is also [tex]\(-\frac{3}{5}\)[/tex].
Both of these lines have the same slope of [tex]\(-\frac{3}{5}\)[/tex].
2. Finding the slope of the perpendicular line:
The slope of a line perpendicular to another is the negative reciprocal of the original slope. Since the slope of the given lines is [tex]\(-\frac{3}{5}\)[/tex], the slope of the line perpendicular to them is:
[tex]\[ m_\text{perpendicular} = \frac{5}{3} \][/tex]
3. Using the point [tex]\((3, 0)\)[/tex] and the perpendicular slope to find the new line:
Using the point-slope form of the equation of a line, [tex]\(y - y_1 = m(x - x_1)\)[/tex], we substitute the given point [tex]\((3,0)\)[/tex] and the perpendicular slope [tex]\(\frac{5}{3}\)[/tex]:
[tex]\[ y - 0 = \frac{5}{3}(x - 3) \][/tex]
[tex]\[ y = \frac{5}{3}x - 5 \][/tex]
This is the equation of the line in slope-intercept form.
4. Converting to standard form Ax + By = C:
[tex]\[ y = \frac{5}{3}x - 5 \][/tex]
Multiplying through by 3 to clear the fraction:
[tex]\[ 3y = 5x - 15 \][/tex]
Rewriting to the standard form:
[tex]\[ 5x - 3y = 15 \][/tex]
5. Checking which of the given options match this line equation:
The given equations are:
- [tex]\(3x + 5y = -9\)[/tex]
- [tex]\(3x + 5y = 9\)[/tex]
- [tex]\(5x - 3y = -15\)[/tex]
- [tex]\(5x - 3y = 15\)[/tex]
From our calculations, we find that the equation that matches is [tex]\(5x - 3y = 15\)[/tex].
Therefore, the equation of the line that is perpendicular to the given lines and passes through the point [tex]\((3, 0)\)[/tex] is:
[tex]\[ \boxed{5x - 3y = 15} \][/tex]
We begin by identifying the slopes of the given lines and then finding the slope of the line that will be perpendicular to them.
1. Identifying the slopes of the given lines:
- The equation [tex]\(3x + 5y = -9\)[/tex] can be rewritten in the slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ 5y = -3x - 9 \][/tex]
[tex]\[ y = -\frac{3}{5}x - \frac{9}{5} \][/tex]
The slope [tex]\(m\)[/tex] is [tex]\(-\frac{3}{5}\)[/tex].
- The equation [tex]\(3x + 5y = 9\)[/tex] can be rewritten in the slope-intercept form [tex]\(y = mx + b\)[/tex]:
[tex]\[ 5y = -3x + 9 \][/tex]
[tex]\[ y = -\frac{3}{5}x + \frac{9}{5} \][/tex]
The slope [tex]\(m\)[/tex] is also [tex]\(-\frac{3}{5}\)[/tex].
Both of these lines have the same slope of [tex]\(-\frac{3}{5}\)[/tex].
2. Finding the slope of the perpendicular line:
The slope of a line perpendicular to another is the negative reciprocal of the original slope. Since the slope of the given lines is [tex]\(-\frac{3}{5}\)[/tex], the slope of the line perpendicular to them is:
[tex]\[ m_\text{perpendicular} = \frac{5}{3} \][/tex]
3. Using the point [tex]\((3, 0)\)[/tex] and the perpendicular slope to find the new line:
Using the point-slope form of the equation of a line, [tex]\(y - y_1 = m(x - x_1)\)[/tex], we substitute the given point [tex]\((3,0)\)[/tex] and the perpendicular slope [tex]\(\frac{5}{3}\)[/tex]:
[tex]\[ y - 0 = \frac{5}{3}(x - 3) \][/tex]
[tex]\[ y = \frac{5}{3}x - 5 \][/tex]
This is the equation of the line in slope-intercept form.
4. Converting to standard form Ax + By = C:
[tex]\[ y = \frac{5}{3}x - 5 \][/tex]
Multiplying through by 3 to clear the fraction:
[tex]\[ 3y = 5x - 15 \][/tex]
Rewriting to the standard form:
[tex]\[ 5x - 3y = 15 \][/tex]
5. Checking which of the given options match this line equation:
The given equations are:
- [tex]\(3x + 5y = -9\)[/tex]
- [tex]\(3x + 5y = 9\)[/tex]
- [tex]\(5x - 3y = -15\)[/tex]
- [tex]\(5x - 3y = 15\)[/tex]
From our calculations, we find that the equation that matches is [tex]\(5x - 3y = 15\)[/tex].
Therefore, the equation of the line that is perpendicular to the given lines and passes through the point [tex]\((3, 0)\)[/tex] is:
[tex]\[ \boxed{5x - 3y = 15} \][/tex]
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