For this case we have that by definition, the equation of the line in a slope-intersection form is given by:
[tex]y = mx + b[/tex]
Where:
m: It's the slope
b: It is the cut-off point with the y axis
We have the following equation:
[tex]y = 2x + 2[/tex]
Thus, the slope is [tex]m_ {1} = 2[/tex]
By definition, if two lines are perpendicular then the product of the slopes is -1.
[tex]m_ {1} * m_ {2} = - 1[/tex]
We find[tex]m_ {2}:[/tex]
[tex]m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {2}[/tex]
Thus, the equation of the line is:
[tex]y = - \frac {1} {2} x + b[/tex]
We substitute the given point to find "b":
[tex]3 = - \frac {1} {2} 6 + b\\3 = -3 + b\\b = 6[/tex]
Thus, the equation of a line perpendicular to the given line and passing through the given point is:
[tex]y = - \frac {1} {2} x + 6[/tex]
Answer:
[tex]y = - \frac {1} {2} x + 6[/tex]
