91Ó°ÊÓ

The slope-intercept form of a linear equation is $y=mx+b$, where $m$is the slope and $b$is the $y$-intercept.

Parallel lines: Slopes $m_1$and $m_2$of the parallel lines are same i.e., $m_1=m_2$.

Perpendicular lines: The slopes $m_1$and $m_2$of the nonvertical perpendicular lines are opposite reciprocals i.e., $m_1{m}_2=−1$.

Calculate the slope of the first equation $−4x+5y=−6$. Convert the given equation in the standard slope and intercept form of the equation $y=mx+b$.

$$\begin{gathered} \text{ }−4x+5y=−6 \\ 5y=4x−6 \\ y=\frac{4}{5}x−\frac{6}{5} \end{gathered}$$

Compared from the standard form of the equation $y=mx+b$, the slope of the equation is:

$m=\frac{4}{5}$

Calculate the slope of the line which is perpendicular to the given line $−4x+5y=−6$.

$$\begin{gathered} m_1{m}_2=−1 \\ \left(\frac{4}{5}\right)m_2=−1 \\ {m}_2=−\frac{5}{4} \end{gathered}$$

To find the equation inslope-intercept form for the line that passes through the point $\left(−4,−5\right)$and is perpendicular to the equation $−4x+5y=−6$,substitute the value of slope and point in the standard form of slope- point form of the equation i.e., $y−y_1=m\left(x−x_1\right)$ where in $x_1=−4,y_1=−5$.

$$\begin{gathered} y−y_1=m\left(x−x_1\right) \\ y−\left(−5\right)=−\frac{5}{4}\left(x−\left(−4\right)\right) \\ y+5=−\frac{5}{4}x−5 \\ y=−\frac{5}{4}x−10 \end{gathered}$$

Therefore, the equation in the slope-intercept form for the line that passes through the point $\left(−4,−5\right)$and is perpendicular to the given equation is $y=−\frac{5}{4}x−10$.