Answer:
The equation of line passes through [tex](4,-1)[/tex] and perpendicular to the graph [tex]y=\frac{7}{2}x-\frac{3}{2}[/tex] is [tex]y=\frac{-2x}{7}+\frac{1}{7}[/tex]
Step-by-step explanation:
Given point is [tex](4,-1)[/tex] and equation of line is [tex]y=\frac{7}{2}x-\frac{3}{2}[/tex]
Let the slope of line that passes through point [tex](4,-1)[/tex] is [tex]m_1[/tex]
And slope of line [tex]y=\frac{7}{2}x-\frac{3}{2}[/tex] is [tex]m_2=\frac{7}{2}[/tex] . As it is in the form of [tex]y=mx+c[/tex]
We know the relation between slope of perpendicular line are given by
[tex]m_1\times m_2=-1\\And\ m_1=\frac{-1}{m_2}[/tex]
So, the slope [tex]m_1=\frac{-1}{\frac{7}{2}}=\frac{-2}{7}[/tex]
Now, we can write the equation of line having point [tex](4,-1)[/tex] and slope [tex]\frac{-2}{7}[/tex]
[tex](y-y_1)=m(x-x_1)\\\\(y-(-1))=\frac{-2}{7}(x-4)\\\\y+1=\frac{-2x}{7}-(\frac{2\times -4}{7})\\ \\y+1=\frac{-2x}{7}+\frac{8}{7}\\\\y=\frac{-2x}{7}+\frac{8}{7}-1\\\\y=\frac{-2x}{7}+\frac{8-7}{7}\\\\y=\frac{-2x}{7}+\frac{1}{7}[/tex]
So, the equation of line passes through [tex](4,-1)[/tex] and perpendicular to the graph [tex]y=\frac{7}{2}x-\frac{3}{2}[/tex] is [tex]y=\frac{-2x}{7}+\frac{1}{7}[/tex]
