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To find the equation of the line that is perpendicular to the given line and passes through the point [tex]\((3,0)\)[/tex], let's work through the solution step-by-step.
Given the equation of the line:
[tex]\[3x + 5y = -9\][/tex]
### Step 1: Determine the slope of the given line
The general form of a line is [tex]\(Ax + By = C\)[/tex]. For the given line:
[tex]\[A = 3, \quad B = 5\][/tex]
The slope [tex]\(m_{\text{given}}\)[/tex] of a line in the form [tex]\(Ax + By = C\)[/tex] is given by:
[tex]\[m_{\text{given}} = -\frac{A}{B} = -\frac{3}{5} = -0.6\][/tex]
### Step 2: Determine the slope of the perpendicular line
The slope of the line perpendicular to another line is the negative reciprocal of the slope of the original line. Therefore, the slope [tex]\(m_{\text{perpendicular}}\)[/tex] is:
[tex]\[m_{\text{perpendicular}} = -\frac{1}{m_{\text{given}}} = -\frac{1}{-0.6} = \frac{5}{3} = 1.6666666666666667\][/tex]
### Step 3: Use point-slope form to find the equation of the perpendicular line
The point-slope form of a line with slope [tex]\(m\)[/tex] through point [tex]\((x_1, y_1)\)[/tex] is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Given point [tex]\((3, 0)\)[/tex]:
[tex]\[ x_1 = 3, \quad y_1 = 0 \][/tex]
Using the slope [tex]\(m = \frac{5}{3}\)[/tex], the equation becomes:
[tex]\[ y - 0 = \frac{5}{3}(x - 3) \][/tex]
[tex]\[ y = \frac{5}{3}x - 5 \][/tex]
### Step 4: Convert the equation to standard form
To express the equation in the form [tex]\(Ax + By = C\)[/tex], multiply both sides by 3 to get rid of the fraction:
[tex]\[ 3y = 5x - 15 \][/tex]
[tex]\[ -5x + 3y = -15 \][/tex]
Multiplying the entire equation by -1 to make it a positive form:
[tex]\[ 5x - 3y = 15 \][/tex]
Thus, the equation of the line that is perpendicular to [tex]\(3x + 5y = -9\)[/tex] and passes through the point [tex]\((3,0)\)[/tex] is:
[tex]\[ \boxed{5x - 3y = 15} \][/tex]
Given the equation of the line:
[tex]\[3x + 5y = -9\][/tex]
### Step 1: Determine the slope of the given line
The general form of a line is [tex]\(Ax + By = C\)[/tex]. For the given line:
[tex]\[A = 3, \quad B = 5\][/tex]
The slope [tex]\(m_{\text{given}}\)[/tex] of a line in the form [tex]\(Ax + By = C\)[/tex] is given by:
[tex]\[m_{\text{given}} = -\frac{A}{B} = -\frac{3}{5} = -0.6\][/tex]
### Step 2: Determine the slope of the perpendicular line
The slope of the line perpendicular to another line is the negative reciprocal of the slope of the original line. Therefore, the slope [tex]\(m_{\text{perpendicular}}\)[/tex] is:
[tex]\[m_{\text{perpendicular}} = -\frac{1}{m_{\text{given}}} = -\frac{1}{-0.6} = \frac{5}{3} = 1.6666666666666667\][/tex]
### Step 3: Use point-slope form to find the equation of the perpendicular line
The point-slope form of a line with slope [tex]\(m\)[/tex] through point [tex]\((x_1, y_1)\)[/tex] is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Given point [tex]\((3, 0)\)[/tex]:
[tex]\[ x_1 = 3, \quad y_1 = 0 \][/tex]
Using the slope [tex]\(m = \frac{5}{3}\)[/tex], the equation becomes:
[tex]\[ y - 0 = \frac{5}{3}(x - 3) \][/tex]
[tex]\[ y = \frac{5}{3}x - 5 \][/tex]
### Step 4: Convert the equation to standard form
To express the equation in the form [tex]\(Ax + By = C\)[/tex], multiply both sides by 3 to get rid of the fraction:
[tex]\[ 3y = 5x - 15 \][/tex]
[tex]\[ -5x + 3y = -15 \][/tex]
Multiplying the entire equation by -1 to make it a positive form:
[tex]\[ 5x - 3y = 15 \][/tex]
Thus, the equation of the line that is perpendicular to [tex]\(3x + 5y = -9\)[/tex] and passes through the point [tex]\((3,0)\)[/tex] is:
[tex]\[ \boxed{5x - 3y = 15} \][/tex]
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