First, we will find the slope of a passing through the given points (4,30) and (1,15).
[tex]\text{Let (x}_1,y_1)=(1,15)and(x_2,y_2)=(4,30)[/tex]
The slope of the line is,
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{30-15}{4-1} \\ m=\frac{15}{3} \\ m=5 \end{gathered}[/tex]
So, the slope of the line is 5. Now, the slope of a line perpendicular to the line with slope m is equal to the negative reciprocal of m. Therefore, the slope of the line perpendicular to the line passing through the points (4,30) and (1,15) is,
[tex]m_1=\frac{-1}{m}=\frac{-1}{5}[/tex]
Let (x1,y1)=(5,0). Let(x,y) be any point on the line passing through the point (5,0) and having slope -1/5.
Now, the equation of this line is,
[tex]\begin{gathered} m_1=\frac{y_1-y}{x_1-x} \\ \frac{-1}{5}=\frac{0-y}{5-x} \\ -1(5-x)=-5y \\ -5+x=-5y \\ x+5y=5 \end{gathered}[/tex]
Therefore, the equation of a line passing through the point (5,0) and perpendicular to a line passing through points (4,30) and (1,15) is x+5y=5. Option C is correct.
