As a first step we are going to calculte the slope of CX:
C(0,3) and X(2, -1)
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{-1-3}{2-0} \\ m=-\frac{4}{2} \\ m=-2 \end{gathered}[/tex]
Slope of a perpendicular line is the negative reciprocal of the original line, in this case the line that passes through T(0, -2) has a slope of -(-1/2)=1/2
So, by the slope-point form of the equation, we can get the equation in slope-intercept form:
[tex]\begin{gathered} y-y_0=m(x-x_0) \\ y+2=\frac{1}{2}(x-0) \\ y=\frac{1}{2}x-2\text{ -> Slope-intercept form} \end{gathered}[/tex]
