Given: The equation below
[tex]4x+7y=4y-7[/tex]To Determine: The equation of the line that passes through the point (- 5, 8) and is perpendicular to the given equation
Solution
Let us determine the slope of the given equation
[tex]\begin{gathered} 4x+7y=4y-7 \\ 7y-4y=-4x-7 \\ 3y=-4x-7 \\ \frac{3y}{3}=\frac{-4x}{3}-\frac{7}{3} \\ y=-\frac{4}{3}x-\frac{7}{3} \end{gathered}[/tex]The slope-intercept form of a linear equation is given as
[tex]\begin{gathered} y=mx+c \\ m=slope \\ c=y-intercept \end{gathered}[/tex]Comparing the slope-intercept form to the given equation
[tex]\begin{gathered} y=-\frac{4}{3}x-\frac{7}{3} \\ y=mx+c \\ slope=m=-\frac{4}{3} \\ c=-\frac{7}{3} \end{gathered}[/tex]Note: If two lines are perpendicular to each other, the slope of one of the line is equal to the negative inverse of the other
Therefore, the slope of the perpendicular line is as shown below
[tex]\begin{gathered} slope(given-equation)=m \\ slope(perpendicular-line)m_2=-(m)^{-1} \\ So \\ m=-\frac{4}{3} \\ m_2=-(-\frac{4}{3})^{-1} \\ m_2=-(-\frac{3}{4}) \\ m_2=\frac{3}{4} \end{gathered}[/tex]If the perpendicular line passes through (-5,8), the equation of the line can be derived using the formula below
[tex]\begin{gathered} \frac{y-y_1}{x-x_1}=slope \\ (x_1,y_1)=(-5,8) \\ slope=\frac{3}{4} \\ Therefore, \\ \frac{y-8}{x--5}=\frac{3}{4} \end{gathered}[/tex][tex]\begin{gathered} \frac{y-8}{x+5}=\frac{3}{4} \\ y-8=\frac{3}{4}(x+5) \\ y-8=\frac{3}{4}x+\frac{15}{4} \\ y=\frac{3}{4}x+\frac{15}{4}+8 \end{gathered}[/tex][tex]\begin{gathered} y=\frac{3}{4}x+\frac{15+32}{4} \\ y=\frac{3}{4}x+\frac{47}{4} \end{gathered}[/tex]Hence, the equation of the line that passes through the point ( - 5, 8) and perpendicular to the given equation is
[tex]y=\frac{3}{4}x+\frac{47}{4}[/tex]
