What is the equation of the line that is perpendicular to the line and passes through the point (6, 2)?

Step [tex]1[/tex]
Find the slope of the line perpendicular to the given line
we have
the equation of the given line is
[tex]y=\frac{1}{2}x+7[/tex]
we know that
if two lines are perpendicular
then
the product of their slopes is equal to minus one
so
[tex]m1*m2=-1[/tex]
in this problem the slope m1 of the given line is
[tex]m1=\frac{1}{2}[/tex]
so
[tex]m2=-\frac{1}{m1}[/tex]
[tex]m2=-2[/tex]
Step [tex]2[/tex]
Find the equation of the line with slope m2 and passes through the point [tex](6, 2)[/tex]
we know that
the equation of the line is
[tex]y-y1=m*(x-x1)\\ y-2=-2*(x-6)\\y=-2x+12+2\\ y=-2x+14[/tex]
therefore
the answer is the option A
the equation of the line is [tex]y=-2x+14[/tex]
see the attached figure to better understand the problem