To find the equation of the line that is perpendicular to the line given by the equation \( 2y + 3x - 1 = 0 \) and passes through the point \( A(2, 4) \), we can follow these steps:
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Rearrange the given line equation to find its slope: The equation \( 2y + 3x - 1 = 0 \) can be rearranged to the slope-intercept form \( y = mx + b \).
\[ 2y = -3x + 1 \] \[ y = -\frac{3}{2}x + \frac{1}{2} \]
Here, the slope (\( m \)) of the line is \( -\frac{3}{2} \).
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Find the slope of the perpendicular line: The slope of a line that is perpendicular to another line is the negative reciprocal of the original slope. Thus, the slope of the perpendicular line is:
\[ m_{\text{perpendicular}} = \frac{2}{3} \]
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Use the point-slope form to find the equation of the perpendicular line: We can now use the point-slope form of the equation of a line, which is given by:
\[ y - y_1 = m(x - x_1) \] where \( (x_1, y_1) \) is the point through which the line passes (in this case, \( (2, 4) \)), and \( m \) is the slope.
Substituting in the values:
\[ y - 4 = \frac{2}{3}(x - 2) \]
Simplifying this:
\[ y - 4 = \frac{2}{3}x - \frac{4}{3} \] \[ y = \frac{2}{3}x - \frac{4}{3} + 4 \] \[ y = \frac{2}{3}x - \frac{4}{3} + \frac{12}{3} \] \[ y = \frac{2}{3}x + \frac{8}{3} \]
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Convert to standard form \( ax + by + c = 0 \): To convert this equation into the required form \( ax + by + c = 0 \), we can multiply through by 3 to eliminate the fractions:
\[ 3y = 2x + 8 \]
Rearranging gives:
\[ -2x + 3y - 8 = 0 \]
Alternatively, we can write it as:
\[ 2x - 3y + 8 = 0 \]
Thus, the equation of the perpendicular line in the form \( ax + by + c = 0 \) is:
\[ \boxed{2x - 3y + 8 = 0} \]