Question

Line v passes through point [6.6] and is perpendicular to the graph of y= 3/4x - 11 line w is parallel to line v and passes through point [-6,10] what is the equation in slope intercept form of line w please help im in a exam

Answer 1

Given:

The equation of a line is

`y=(3)/(4)x-11`

Line v passes through point (6,6) and it is perpendicular to the given line.

Line w passes through point (-6,10) and it is parallel to the line v.

To find:

The equation in slope intercept form of line w.

Solution:

Slope intercept form of a line is

`y=mx+b` ...(i)

where, m is slope and b is y-intercept.

We have,

`y=(3)/(4)x-11` ...(ii)

On comparing (i) and (ii), we get

`m=(3)/(4)`

So, slope of given line is
`(3)/(4)`.

Product of slopes of two perpendicular lines is -1.

`m_1* m_2=-1`

`(3)/(4)* m_2=-1`

`m_2=-(4)/(3)`

Line w is perpendicular to the given line. So, the slope of line w is
`-(4)/(3)`.

Slopes of parallel line are equal.

Line v is parallel to line w. So, slope of line v is also
`-(4)/(3)`.

Slope of line v is
`-(4)/(3)` and it passes thorugh (-6,10). So, the equation of line v is

`y-y_1=m(x-x_1)`

where, m is slope.

`y-10=-(4)/(3)(x-(-6))`

`y-10=-(4)/(3)(x+6)`

`y-10=-(4)/(3)x-(4)/(3)(6)`

`y-10=-(4)/(3)x-8`

Adding 10 on both sides, we get

`y=-(4)/(3)x-8+10`

`y=-(4)/(3)x+2`

Therefore the equation of line v is
`y=-(4)/(3)x+2`.