Question

Write the equation in slope intercept form of a line that is perpendicular to -2x+4y=12 and passes through (-8,1)

Answer 1

The equation of the line is given below as

`-2x+4y=12`

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

`m_1* m_2=-1`

Step 1: make y the subject of the formula in the equation of the line below

`-2x+4y=12`

Add 2x to both sides of the equation

`\begin{gathered} -2x+2x+4y=12+2x \\ 4y=2x+12 \end{gathered}`

Divide all through by 4

`\begin{gathered} 4y=2x+12 \\ (4y)/(4)=(2x)/(4)+(12)/(4) \\ y=(1)/(2)x+3 \end{gathered}`

The general equation of a line is

`\begin{gathered} y=mx+c \\ \text{where,} \\ m=\text{slope} \\ c=y-\text{intercept} \end{gathered}`

By comparing coefficients,

From the equation above, the slope m1 is

`m_1=(1)/(2)`

Recall that for a perpendicular line,

`\begin{gathered} m_1* m_2=-1 \\ (1)/(2)* m_2=-1 \\ (m_2)/(2)=-1 \\ m_2=-1*2 \\ m_2=-2 \end{gathered}`

Step 2: Calculate the equation of the perpendicular line using the formula below

`\begin{gathered} m_2=(y-y_1)/(x-x_1) \\ \text{where,} \\ x_1=-8,y_1=1 \end{gathered}`

By substituting the values, we will have

`\begin{gathered} m_2=(y-y_1)/(x-x_1) \\ -2=(y-1)/(x-(-8)) \\ -2=(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}`

Hence,

The equation of the line in slope-intercept form is

y = -2x -15