Answer :
Recall that the equation of a line in slope-intercept form is given by
[tex]y=mx+b[/tex]Where m is the slope and b is the y-intercept.
Let us re-write the given equations into the slope-intercept form to identify their slopes.
[tex]\begin{gathered} -6x+8y=2 \\ 8y=6x+2 \\ y=\frac{6x}{8}+\frac{2}{8} \\ y=\frac{3x}{4}+\frac{1}{4} \end{gathered}[/tex]So, the slope of the 1st equation is 3/4
[tex]m_1=\frac{3}{4}[/tex]Now, the 2nd equation becomes,
[tex]\begin{gathered} -8x+6y=1 \\ 6y=8x+1 \\ y=\frac{8x}{6}+\frac{1}{6} \\ y=\frac{4x}{3}+\frac{1}{6} \end{gathered}[/tex]So, the slope of the 2nd equation is 4/3
[tex]m_2=\frac{4}{3}[/tex]Recall that two equations are perpendicular if their slopes are negative reciprocal of each other.
Mathematically,
[tex]m_2=-\frac{1}{m_1}[/tex]Substitute the values of the slopes and check if the relation holds true.
[tex]\begin{gathered} \frac{4}{3}=-\frac{1}{\frac{3}{4}} \\ \frac{4}{3}\ne-\frac{4}{3} \end{gathered}[/tex]As you can see, the relation does not hold true since one is positive and the other is negative.
Therefore, we can conclude that the given equations are not perpendicular.
