The equation of the line is given as,
[tex]y=2x+2[/tex]
According to the slope-intercept form, the equation of a line with slope 'm' and y-intercept 'c', is given by,
[tex]y=mx+c[/tex]
Comparing with the given equation, the slope of the given line is,
[tex]m=2[/tex]
Theorem: The product of two slopes of two perpendicular lines is -1 always.
Let the slope of the perpendicular line be (m'), and 'b' be the y-intercept. Then, it follows that,
[tex]\begin{gathered} m^{\prime}\cdot m=-1 \\ m^{\prime}\cdot(2)=-1 \\ m^{\prime}=\frac{-1}{2} \end{gathered}[/tex]
The equation of this perpendicular line is given by,
[tex]\begin{gathered} y=m^{\prime}x+b^{\prime} \\ y=(\frac{-1}{2})x+b^{\prime} \end{gathered}[/tex]
Given that the perpendicular lines pass through the point (6,-2), so it must also satisfy its equation,
[tex]\begin{gathered} -2=(\frac{-1}{2})(6)+b^{\prime} \\ -2=-3+b^{\prime} \\ b^{\prime}=-2+3 \\ b^{\prime}=1 \end{gathered}[/tex]
Substitute this value back in the equation of the perpendicular line,
[tex]y=\frac{-1}{2}x+1[/tex]
Thus, the equation of a line perpendicular to the given line and passing through the given point is obtained as,
[tex]y=\frac{-1}{2}x+1[/tex]
