SOLUTION
For a line to be perpendicular to another line, product of thier gradient must be -1.
if the gradient is given as
[tex]\begin{gathered} m_{1\text{ }}andm_{2\text{ }},\text{ then} \\ m_1m_{2=-1} \end{gathered}[/tex]Then given the line
[tex]y=\frac{6}{7}x+2[/tex]The gradient of the line is the coefficient of x using the expression
[tex]\begin{gathered} y=mx+c \\ m=Gradient \end{gathered}[/tex]Hence, we have
[tex]m_1=\frac{6}{7}[/tex]Then, using
[tex]\begin{gathered} m_1m_2=-1 \\ m_2=\frac{-1}{m_1} \\ m_2=-1\times\frac{7}{6}=-\frac{7}{6} \end{gathered}[/tex]Given the point (-6,4), the line perpendicular will be having the equation
[tex]\begin{gathered} y-y_1=m_2(x-x_1) \\ \text{where }y_1=4\text{ and x}_1=-6 \end{gathered}[/tex]Then we obtain
[tex]\begin{gathered} y-4=-\frac{7}{6}(x-(-6)) \\ y-4=-\frac{7}{6}x-7 \\ y=-\frac{7}{6}x-7+4 \\ y=-\frac{7}{6}x-3 \end{gathered}[/tex]Therefore the equation perpendicular to this line is given as y=-7/6x - 3
[tex]y=-\frac{7}{6}x-3[/tex]Then the equation parallel to the same line will have the same gradient
[tex]m_1=m_2=\frac{6}{7}[/tex]Then the equation parallel passing through the point (-6,4) will be
[tex]\begin{gathered} y-4=\frac{6}{7}(x-(-6)) \\ y-4=\frac{6}{7}(x+6) \\ y-4=\frac{6}{7}x+\frac{36}{7} \\ y=\frac{6}{7}x+\frac{36}{7}+4 \\ y=\frac{6}{7}x+\frac{64}{7} \end{gathered}[/tex]The equation of the line that is parallel to this line and passes through the point (-6, 4) will be y=6/7x+64/7
[tex]y=\frac{6}{7}x+\frac{64}{7}[/tex]
