[tex]y=-\frac{1}{2}x+\frac{3}{2}[/tex]
1) In this question, let's find the equation, using the point-slope formula:
[tex](y-y_0)=m(x-x_0)[/tex]
2) Notice that since we want a perpendicular line we can write a perpendicular line to y=2, as x=-1/2 for -1/2 is the opposite and reciprocal to 2 (the necessary condition to get a perpendicular line).
So, the slope of that perpendicular line is -1/2
3) Let's plug into that Point-Slope formula, the slope m= -1/2 and the point:
[tex]\begin{gathered} (y-2)=-\frac{1}{2}(x+1) \\ y-2=-\frac{1}{2}x-\frac{1}{2} \\ y=-\frac{1}{2}x-\frac{1}{2}+2 \\ y=-\frac{1}{2}x+\frac{3}{2} \end{gathered}[/tex]
4) Thus, the answer is:
[tex]y=-\frac{1}{2}x+\frac{3}{2}[/tex]
