Given: The equation below
[tex]\begin{gathered} y=2x+4 \\ Point:(2,-2) \end{gathered}[/tex]
To Determine: The equation that is perpendicular to the given equation
Solution
The given equation can be represented using the slope-intercept form
[tex]\begin{gathered} slope-intercept-form:y=mx+c \\ Where \\ m=slope \\ c=intercept \end{gathered}[/tex]
Let us determine the slope of the equation
[tex]\begin{gathered} y=mx+c \\ y=2x+4 \\ m=2,c=4 \end{gathered}[/tex]
Therefore, the slope is 2
Please note two linear equations are perpedicular if the slope one is a negative inverse of the other
So, we have
[tex]\begin{gathered} m_1=-\frac{1}{m} \\ m_1=slope\text{ of the perpendicular equation} \end{gathered}[/tex]
Given the slope and a point, we can determine the equation using the formula below
[tex]\begin{gathered} point(x_1,y_1),slope(m) \\ \frac{y-y_1}{x-x_1}=m \end{gathered}[/tex]
Let us substitute the slope and the coordinate of the points given
[tex]\begin{gathered} \frac{y--2}{x-2}=-\frac{1}{2} \\ \frac{y+2}{x-2}=-\frac{1}{2} \\ y+2=-\frac{1}{2}(x-2) \\ y+2=-\frac{1}{2}x+1 \\ y=-\frac{1}{2}x+1-2 \\ y=-\frac{1}{2}x-1 \end{gathered}[/tex]
Hence, the equation of the perpendicular to the given equation is
[tex]y=-\frac{1}{2}x-1[/tex]
