Consider that the standard equation of a line with slope (m) and y-intercept (c) is given by,
[tex]y=mx+c[/tex]Comparing with the given equation, the slope is obtained as,
[tex]m=-4[/tex]Let m' be the slope of the perpendicular line.
The line will be perpendicular only if their product of slopes is -1,
[tex]\begin{gathered} m\times m^{\prime}=-1 \\ -4\times m^{\prime}=-1 \\ m^{\prime}=\frac{-1}{-4} \\ m^{\prime}=\frac{1}{4} \end{gathered}[/tex]So the equation of this perpendicular line is given by,
[tex]y=\frac{1}{4}x+c^{\prime}[/tex]Since the perpendicular if at point (4,10), so it must satisfy its equation,
[tex]\begin{gathered} 10=\frac{1}{4}(4)+c^{\prime} \\ 10=1+c^{\prime} \\ c^{\prime}=9 \end{gathered}[/tex]Substitute the value back in the equation,
[tex]\begin{gathered} y=\frac{1}{4}x+9 \\ 4y=x+36 \\ x-4y+36=0 \end{gathered}[/tex]Thus, the required equation which is perpendicular to the given line is x-4y+36=0.
