Respuesta :

Solution:

Given:

[tex]\begin{gathered} Line\text{ 1:} \\ -x-5y=7 \\ \\ Line\text{ 2:} \\ 5x-y=-9 \end{gathered}[/tex]

Rewriting the equations of the lines given in the slope-intercept form to get the slopes of the lines;

[tex]\begin{gathered} y=mx+b \\ where: \\ m\text{ is the slope} \\ b\text{ is the y-intercept} \\ \\ Then,\text{ the lines can be rewritten as;} \\ Line\text{ 1:} \\ -x-5y=7 \\ -5y=7+x \\ -5y=x+7 \\ y=\frac{x+7}{-5} \\ Splitting\text{ the terms;} \\ y=-\frac{1}{5}x-\frac{7}{5} \\ Hence,\text{ the slope of line 1 is;} \\ m_1=-\frac{1}{5} \end{gathered}[/tex]

Also, for line 2

[tex]\begin{gathered} Rewriting\text{ line 2;} \\ Line\text{ 2:} \\ 5x-y=-9 \\ 5x+9=y \\ y=5x+9 \\ Then,\text{ the slope of line 2 is;} \\ m_2=5 \end{gathered}[/tex]

Two lines are perpendicular when the product of their slopes is -1.

Hence,

[tex]If\text{ }m_1m_2=-1,\text{ then the lines are perpendicular}[/tex]

Multiplying the slopes of the two lines gotten;

[tex]\begin{gathered} m_1m_2=-\frac{1}{5}\times5 \\ m_1m_2=-1 \end{gathered}[/tex]

Since the product of their slopes is -1, then the two lines are perpendicular.

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