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Sagot :
if two lines are perpendicular, it is true that:
[tex]\begin{gathered} m1\cdot m2=-1 \\ Let\colon \\ m1=\frac{4}{7} \\ m2=other_{\text{ }}line \\ \frac{4}{7}\cdot m2=-1 \\ solve_{\text{ }}for_{\text{ }}m2 \\ m2=-1\cdot\frac{7}{4} \\ m2=-\frac{7}{4} \end{gathered}[/tex]
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