Answer :
We know that the product of to slope of perpendicular lines is always equal to 1, so for a) we have that
[tex]\begin{gathered} \frac{4}{3}\cdot m^{\prime}\text{ = -1} \\ m^{\prime}\text{ = (-1)}\frac{3}{4} \\ m^{\prime}\text{ = }\frac{-3}{4} \end{gathered}[/tex]The answer is: the slope of the line perperdincular is m' = -3/4.
Now for b) we have that
[tex]\begin{gathered} -\frac{3}{7}\cdot\text{ m' = -1} \\ m^{\prime}\text{ = (-1)}\cdot(-\frac{7}{3}) \\ m^{\prime}\text{ = }\frac{7}{3} \end{gathered}[/tex]The answer is: the slope of the line perperdincular is m' = 7/3.
Now for c) we have that
[tex]\begin{gathered} 4\cdot m^{\prime}\text{ = -1} \\ m^{\prime}\text{ = }\frac{-1}{4} \end{gathered}[/tex]The answer is: the slope of the line perperdincular is m' = -1/4.
Now for d) we have that
[tex]\begin{gathered} -\frac{1}{3}\cdot m^{\prime}\text{ = -1} \\ m^{\prime}\text{ = (-1)}\frac{-3}{1}\text{ } \\ m^{\prime}\text{ = 3} \end{gathered}[/tex]The answer is: the slope of the line perperdincular is m' = 3.
Now for e) we have that
[tex]\begin{gathered} 1\cdot\text{ m' = -1} \\ m^{\prime}\text{ = }\frac{-1}{1} \\ m^{\prime}\text{ = -1} \end{gathered}[/tex]The answer is: the slope of the line perperdincular is m' = -1.
