Question

PLEASE ANSWER THIS ASAP

Write an equation in slope intercept form for the line that passes through (4,-1) and is perpendicular to the graph of y=7/2x-3/2

Answer 1

The equation of line passes through
`(4,-1)` and perpendicular to the graph
`y=(7)/(2)x-(3)/(2)` is
`y=(-2x)/(7)+(1)/(7)`

Explanation:

Given point is
`(4,-1)` and equation of line is
`y=(7)/(2)x-(3)/(2)`

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

And slope of line
`y=(7)/(2)x-(3)/(2)` is
`m_2=(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* m_2=-1\\And\ m_1=(-1)/(m_2)`

So, the slope
`m_1=(-1)/((7)/(2))=(-2)/(7)`

Now, we can write the equation of line having point
`(4,-1)` and slope
`(-2)/(7)`

`(y-y_1)=m(x-x_1)\\\\(y-(-1))=(-2)/(7)(x-4)\\\\y+1=(-2x)/(7)-((2* -4)/(7))\\ \\y+1=(-2x)/(7)+(8)/(7)\\\\y=(-2x)/(7)+(8)/(7)-1\\\\y=(-2x)/(7)+(8-7)/(7)\\\\y=(-2x)/(7)+(1)/(7)`

So, the equation of line passes through
`(4,-1)` and perpendicular to the graph
`y=(7)/(2)x-(3)/(2)` is
`y=(-2x)/(7)+(1)/(7)`