Question

PLEASE ANSWER THIS ASAP

Answer 1

Answer:

The equation of line passes through $(4,-1)$ and perpendicular to the graph $y=\frac{7}{2}x-\frac{3}{2}$ is $y=\frac{-2x}{7}+\frac{1}{7}$

Step-by-step explanation:

Given point is $(4,-1)$ and equation of line is $y=\frac{7}{2}x-\frac{3}{2}$

Let the slope of line that passes through point $(4,-1)$ is $m_1$

And slope of line $y=\frac{7}{2}x-\frac{3}{2}$ is $m_2=\frac{7}{2}$ . As it is in the form of $y=mx+c$

We know the relation between slope of perpendicular line are given by

$m_1\times m_2=-1\\And\ m_1=\frac{-1}{m_2}$

So, the slope $m_1=\frac{-1}{\frac{7}{2}}=\frac{-2}{7}$

Now, we can write the equation of line having point  $(4,-1)$ and slope $\frac{-2}{7}$

$(y-y_1)=m(x-x_1)\\\\(y-(-1))=\frac{-2}{7}(x-4)\\\\y+1=\frac{-2x}{7}-(\frac{2\times -4}{7})\\ \\y+1=\frac{-2x}{7}+\frac{8}{7}\\\\y=\frac{-2x}{7}+\frac{8}{7}-1\\\\y=\frac{-2x}{7}+\frac{8-7}{7}\\\\y=\frac{-2x}{7}+\frac{1}{7}$

So, the equation of line passes through $(4,-1)$ and perpendicular to the graph $y=\frac{7}{2}x-\frac{3}{2}$ is $y=\frac{-2x}{7}+\frac{1}{7}$