Find an equation of the line perpendicular to \( x+3 y=-5 \) and passing through \( (-13,-1) \). Express the equation in standard form.
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To find the equation of a line that is perpendicular to the given line \( x + 3y = -5 \), we first need to determine the slope of the given line. Rewriting it in slope-intercept form \( y = mx + b \): \[ 3y = -x - 5 \implies y = -\frac{1}{3}x - \frac{5}{3} \] The slope of this line is \( -\frac{1}{3} \). The slope of the line perpendicular to it will be the negative reciprocal, which is \( 3 \). Now, we use the point-slope form of a line, which is given by \( y - y_1 = m(x - x_1) \). Given the point \( (-13, -1) \) and the slope \( 3 \): \[ y - (-1) = 3(x - (-13)) \implies y + 1 = 3(x + 13) \] Expanding this: \[ y + 1 = 3x + 39 \implies y = 3x + 38 \] Now, we convert this to standard form \( Ax + By = C \). We move \( y \) to the left side: \[ -3x + y = 38 \] To write it in the standard form where \( A \) is positive, we can multiply through by \( -1 \): \[ 3x - y = -38 \] Thus, the equation of the line in standard form is: \[ \boxed{3x - y = -38} \]
