The equation of the line is given below as
[tex]-2x+4y=12[/tex]
Concept: To get the equation of the line perpendicular to the line above, we will have to calculate the slope of the line using the formula below
[tex]m_1\times m_2=-1[/tex]
Step 1: make y the subject of the formula in the equation of the line below
[tex]-2x+4y=12[/tex]
Add 2x to both sides of the equation
[tex]\begin{gathered} -2x+2x+4y=12+2x \\ 4y=2x+12 \end{gathered}[/tex]
Divide all through by 4
[tex]\begin{gathered} 4y=2x+12 \\ \frac{4y}{4}=\frac{2x}{4}+\frac{12}{4} \\ y=\frac{1}{2}x+3 \end{gathered}[/tex]
The general equation of a line is
[tex]\begin{gathered} y=mx+c \\ \text{where,} \\ m=\text{slope} \\ c=y-\text{intercept} \end{gathered}[/tex]
By comparing coefficients,
From the equation above, the slope m1 is
[tex]m_1=\frac{1}{2}[/tex]
Recall that for a perpendicular line,
[tex]\begin{gathered} m_1\times m_2=-1 \\ \frac{1}{2}\times m_2=-1 \\ \frac{m_2}{2}=-1 \\ m_2=-1\times2 \\ m_2=-2 \end{gathered}[/tex]
Step 2: Calculate the equation of the perpendicular line using the formula below
[tex]\begin{gathered} m_2=\frac{y-y_1}{x-x_1} \\ \text{where,} \\ x_1=-8,y_1=1 \end{gathered}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} m_2=\frac{y-y_1}{x-x_1} \\ -2=\frac{y-1}{x-(-8)} \\ -2=\frac{y-1}{x+8} \\ y-1=-2(x+8) \\ y-1=-2x-16 \\ \text{add 1 to both sides} \\ y-1+1=-2x-16+1 \\ y=-2x-15 \end{gathered}[/tex]
Hence,
The equation of the line in slope-intercept form is
y = -2x -15
