[tex]\boxed{y=-5x+21}[/tex]
In this exercise, we have to find the equation of the line that is perpendicular to:
[tex]y=\frac{1}{5}x+4[/tex]
The key point here is that the slopes of perpendicular lines are opposite reciprocals. If we define the slope of the given line as [tex]m[/tex] and the slope of the line we are looking for as [tex]m_{x}[/tex], then:
[tex]m\cdotpm_{x}=-1 \\ \\ Solving \ for \ m_{x}: \\ \\ m_{x}=-\frac{1}{m} \\ \\ But \ m=\frac{1}{5} \\ \\ m_{x}=-5[/tex]
Another information is that the line passes through (5, -4), so if we write the line in Slope-Intercept form, we get:
[tex]y=m_{x}x+b \\ \\ y=-5x+b \\ \\ Goal \rightarrow b \ (y-intercept) \\ \\ \\ For \ x=5 \ and y=-4 \\ \\ -4=-5(5)+b \\ \\ \\ Isolating \ b: \\ \\ b=-4+25 \\ \\ b=21[/tex]
Finally, the What is the equation of the line that is perpendicular to y=1/5x+4 and that passes through (5,-4) is y = -5x + 21
Parallel lines: https://brainly.com/question/12169569
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