Question

[30 POINTS] Please help!!!

Answer 1

Part 1)
`y=1.5x+5`

Part 2)
`y=-(2/3)x-(11/3)`

Part 3)
`y=0.25x+2.75`

Part 4)
`y=-2x+5`

Part 5)
`y=0.5x-1`

Part 6) The graph in the attached figure

Explanation:

Part 1) we have

`m=3/2=1.5`

`point(-2,2)`

The equation of the line into point slope form is equal to

`y-y1=m(x-x1)`

substitute

`y-2=1.5(x+2)`

`y=1.5x+3+2`

`y=1.5x+5`

Part 2) we know that

If two lines are perpendicular

then

the product of their slopes is equal to minus one

so

`m1*m2=-1`

the slope of the line 1 is equal to

`m1=1.5`

Find the slope m2

`1.5*m2=-1`

`m2=-2/3`

Find the equation of the line 2

we have

`m2=-2/3`

`point(-7,1)`

The equation of the line into point slope form is equal to

`y-y1=m(x-x1)`

substitute

`y-1=(-2/3)(x+7)`

`y=-(2/3)x-(14/3)+1`

`y=-(2/3)x-(11/3)`

Part 3) we have

`m=1/4=0.25`

`point(1,3)`

The equation of the line into point slope form is equal to

`y-y1=m(x-x1)`

substitute

`y-3=0.25(x-1)`

`y=0.25x-0.25+3`

`y=0.25x+2.75`

Part 4) we have

`m=-2`

`b=5` -----> y-intercept

we know that

The equation of the line into slope intercept form is equal to

`y=mx+b`

substitute the values

`y=-2x+5`

Part 5) we have that

The slope of the line 4 is equal to
`-2`

so

the slope of the line perpendicular to the line 4 is equal to

`-2*m=-1\\m=(1/2)=0.5`

therefore

in this problem we have

`m=0.5`

`point(-2,-2)`

The equation of the line into point slope form is equal to

`y-y1=m(x-x1)`

substitute

`y+2=0.5(x+2)`

`y=0.5x+1-2`

`y=0.5x-1`

Part 6)

using a graphing tool

see the attached figure