What is an equation of the line that passes through the point \( (4,-6) \) and is perpendicular to the line \( 2 x-3 y=27 \) ?
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To find the equation of the line that passes through the point \( (4,-6) \) and is perpendicular to the line \( 2x - 3y = 27 \), we first need to determine the slope of the given line. We can rearrange the equation into slope-intercept form \( y = mx + b \). Starting with \( 2x - 3y = 27 \), we isolate \( y \): \[ -3y = -2x + 27 \] \[ y = \frac{2}{3}x - 9 \] This means the slope \( m_1 \) of the original line is \( \frac{2}{3} \). For a line to be perpendicular, its slope \( m_2 \) will be the negative reciprocal of \( m_1 \): \[ m_2 = -\frac{1}{(\frac{2}{3})} = -\frac{3}{2} \] Now we have the slope of the line we want to find, which is \( -\frac{3}{2} \). We can use the point-slope form of a line, given by \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point \( (4, -6) \): \[ y - (-6) = -\frac{3}{2}(x - 4) \] \[ y + 6 = -\frac{3}{2}x + 6 \] Subtracting 6 from both sides gives: \[ y = -\frac{3}{2}x + 0 \] Therefore, the equation of the line is: \[ y = -\frac{3}{2}x \]
