Respuesta :
Consider that the equation of a line with slope 'm' and y-intercept 'c' is given by,
[tex]y=mx+c[/tex]Consider the given equation of line,
[tex]\begin{gathered} y-3x=8 \\ y=3x+8 \end{gathered}[/tex]Comparing the coefficient, it is found that the slope of the given line is 3,
[tex]m=3[/tex]Let 's' be the slope of the line which is perpendicular to this line.
Consider that two lines will be perpendicular if their product of slopes is -1,
[tex]\begin{gathered} m\times s=-1 \\ 3\times s=-1 \\ s=\frac{-1}{3} \end{gathered}[/tex]So the slope of the perpendicular line is given by,
[tex]y=\frac{-1}{3}x+c[/tex]Now, it is given that this line passes through the point (-2,5), so it must satisfy the equation of the line,
[tex]\begin{gathered} 5=\frac{-1}{3}(-2)+c_{} \\ 5=\frac{2}{3}+c \\ c=5-\frac{2}{3} \\ c=\frac{13}{3} \end{gathered}[/tex]Substitute the value of 'c' to get the final equation,
[tex]\begin{gathered} y=\frac{-1}{3}x+\frac{13}{3} \\ 3y=-x+13 \\ x+3y=13 \end{gathered}[/tex]Thus, the required equation of the perpendicular line is x + 3y = 13 .
