The line
[tex]10x+5y=10[/tex]Can be written as:
[tex]\begin{gathered} 10x+5y=10 \\ 5y=-10x+10 \\ y=-\frac{10}{5}x+\frac{10}{5} \\ y=-2x+2 \end{gathered}[/tex]This is the same line in the slope-intercept form:
[tex]y=mx+b[/tex]Then, the original line has an slope of -2.
To find the slope of a perpendicular line to this one we need to remmeber that two lines are perpendicular if and only if
[tex]m_1m_2=-1_{}_{}[/tex]Plugging the value of the original line we have:
[tex]\begin{gathered} -2m_2=-1 \\ m_2=-\frac{1}{-2} \\ m_2=\frac{1}{2} \end{gathered}[/tex]Therefore a line perpendicular to the line 10x+5y=10 has slope 1/2 or 0.5.
