Find an equation of the line that contains the point $( -1,3)$ and is perpendicular to the line $4x+y=-1.$

Solution  We begin by writing the equation of the given line in slope-intercept form to find its slope. $$\begin{eqnarray*} 4x+y &=&-1 \\[0pt] y &=&-4x-1 \end{eqnarray*}$$

Figure 32

This line has a slope of $-4.$ Any line perpendicular to this line will have slope $\dfrac{1}{4}.$ Because the point $( -1,3)$ is on this line, we use the point-slope form of a line. $$\begin{eqnarray*} y-y_{1} &=&m( x-x_{1}) \\[3pt] y-3 &=&\dfrac{1}{4} [ x- ( -1) ] \\[4pt] y-3 &=&\dfrac{1}{4}x+\dfrac{1}{4} \\[4pt] y &=&\dfrac{1}{4}x+\dfrac{13}{4} \end{eqnarray*}$$

Figure 32 shows the graphs of $4x+y=-1$ and $y=\dfrac{1}{4}x+\dfrac{13}{4}$.