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Sagot :
Comparing y=3x+8 with straight line equation y=mx+c, where m is the slope, we get
Slope of line y=3x+8, m=3.
The slope of a line perpendicular to y=3x+8 is,
[tex]\begin{gathered} m_1=\frac{-1}{m} \\ =\frac{-1}{3} \end{gathered}[/tex]The coordinates of a point passing through perpendicular line to y=3x+8 is (x0,y0)=(-4,1).
Now, use point slope formula to find the equation of a line passing through (x0,y0)=(-4,1) with slope m1.
[tex]\begin{gathered} m_1=\frac{y-y_0}{x-x_0} \\ \frac{-1}{3}=\frac{y-1}{x-(-4)} \\ \frac{-1}{3}=\frac{y-1}{x+4} \\ -x-4=3y-3 \\ -x-1=3y \\ \end{gathered}[/tex]Therefore, the equation of a line perpendicular to the line y=3x +8 and passes through (-4,1) is -x-1=3y.
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