The genral equation of straight line is y=mx+c, where m is the slope of the line and c is the y intercept.
The given line is x+3y=15. Rewrite this equation in the form y=mx+c.
[tex]\begin{gathered} x+3y=15 \\ 3y=-x+15 \\ y=\frac{-1}{3}x+\frac{15}{3} \\ y=\frac{-1}{3}x+5 \end{gathered}[/tex]
Comparing above equation with y=mx+c, we can write
[tex]\text{Slope, m=}\frac{-1}{3}[/tex]
The slope of a line perpendicular to x+3y=15 is the negative reciprocal of the slope m. Hence, the slope of the perpendicular line can be written as,
[tex]m_1=-\frac{1}{m}=-(\frac{1}{\frac{-1}{3}})=3[/tex]
Let (x1,y1)=( -2,5). Now, the equation of the line with slope m1 and passing through (x1,y1) can be written as,
[tex]\begin{gathered} m_1=\frac{y_1-y}{x_1-x} \\ 3=\frac{5-y}{-2-x} \\ 3(-2-x)=5-y \\ -6-3x=5-y \\ y=3x+11 \end{gathered}[/tex]
Theefore, the equation of aline passing through (-2,5) and perpendicular to x+3y=15 is y=3x+11.
