Write the equation of a line that is perpendicular to the given line and that passes through the given point.–x + 5y = 14; (–5, –2)
Accepted Solution
A:
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1) Perpendicular to [tex]-x+5y=14[/tex] In order for lines to be perpendicular, their slopes must be negative reciprocals. Example of slopes with negative reciprocals: 5 and [tex] \frac{-1}{5} [/tex]
First, rearrange the equation into slope y-intercept form: [tex]-x+5y=14[/tex]
[tex]5y=x+14[/tex]
[tex] \frac{5y}{5}=\frac{x+14}{5} [/tex]
[tex]y=\frac{1}{5}x+\frac{14}{5}[/tex]
The slope of the equation is: \frac{1}{5}
The negative reciprocal formula: [tex] (m_{1})(m_{2})=-1[/tex]
Solve for the negative reciprocal: [tex] \frac{1}{5}m_{2}=-1[/tex]
Divide both sides by [tex]\frac{1}{5}[/tex] [tex]\frac{\frac{1}{5}m_{2} }{ \frac{1}{5}} = \frac{-1}{ \frac{1}{5}}[/tex]
[tex]m_{2}=(-1)(\frac{5}{1})[/tex]
[tex]m_{2}=(\frac{-5}{1})[/tex]
[tex]m_{2}=-5[/tex]
The slope of the new line is: -5
2) Passes through (-5,-2)
Create an equation with the slope discovered in slope y-intercept form. [tex]y=-5x+b[/tex]
Input the point the line passes through. [tex](-2)=-5(-5)+b[/tex]
Solve for b (the y-intercept). [tex]-2=-5(-5)+b[/tex]