Question

HELPPPP!!!!!!!! 100 points!!!! I need help on 37-39 thanks!

Answer 1

Part 37)

a) Point-slope form
`y-5=(1)(x-3)`

b) Slope intercept form
`y=x+2`

c) Standard form
`x-y=-2`

Part 38)

a) Point-slope form
`y-2=0.25(x+4)`

b) Slope intercept form
`y=0.25x+3`

c) Standard form
`x-4y=-12`

Part 39)

a) Point-slope form
`y-3=2(x-1)`

b) Slope intercept form
`y=2x+1`

c) Standard form
`2x-y=-1`

Explanation:

Part 37) Line passing through point (3,5) and slope of 1

Part a) Equation of the line in point slope form

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

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

substitute the given values

`y-5=(1)(x-3)`

`y-5=(x-3)` -----> equation of the line in point slope form

Part b) Equation of the line into slope intercept form

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

`y=mx+b`

where

m is the slope

b is the y-intercept

Substitute the given values in the equation and solve for b

`5=(1)(3)+b`

`5=3+b`

`b=5-3`

`b=2`

substitute

`y=x+2` -----> equation of the line in slope intercept form

Part c) Equation of the line in standard form

The equation of the line in standard form is equal to

`Ax+By=C`

where

A is a positive integer

B and C are integers

we have

`y=x+2`

Convert to standard form

Subtract y both sides

`y-y=x+2-y`

`0=x+2-y`

Subtract 2 both sides

`-2=x+2-y-2`

`-2=x-y`

Rewrite

`x-y=-2` ----> equation in standard form

Part 38) Line passing through points (-4,2) and (0,3)

Part a) Equation of the line in point slope form

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

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

Find the slope m

`m=(3-2)/(0+4)=1/4=0.25`

with the slope m=0.25 and point (-4,2)

substitute the given values

`y-2=0.25(x+4)` -----> equation of the line in point slope form

Part b) Equation of the line into slope intercept form

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

`y=mx+b`

where

m is the slope

b is the y-intercept

we have

`m=0.25`

`b=3` -----> the y-intercept is the point (0,3)

Substitute the given values in the equation

`y=0.25x+3` -----> equation of the line in slope intercept form

Part c) Equation of the line in standard form

The equation of the line in standard form is equal to

`Ax+By=C`

where

A is a positive integer

B and C are integers

we have

`y=0.25x+3`

Convert to standard form

Multiply by 4 both sides to remove the decimal number

`4y=x+12`

Subtract 4y both sides

`4y-4y=x+12-4y`

`0=x+12-4y`

Subtract 12 both sides

`-12=x+12-4y-12`

`-12=x-4y`

Rewrite

`x-4y=-12` ----> equation in standard form

Part 39) Line passing through points (1,3) and (2,5)

Part a) Equation of the line in point slope form

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

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

Find the slope m

`m=(5-3)/(2-1)=2/1=2`

with the slope m=2 and point (1,3)

substitute the given values

`y-3=2(x-1)` -----> equation of the line in point slope form

Part b) Equation of the line into slope intercept form

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

`y=mx+b`

where

m is the slope

b is the y-intercept

with the slope m=2 and point (1,3)

Substitute the given values and solve for b

`3=(2)(1)+b`

`3=2+b`

`b=3-2=1`

substitute

`y=2x+1` -----> equation of the line in slope intercept form

Part c) Equation of the line in standard form

The equation of the line in standard form is equal to

`Ax+By=C`

where

A is a positive integer

B and C are integers

we have

`y=2x+1`

Convert to standard form

Subtract y both sides

`y-y=2x+1-y`

`0=2x+1-y`

Subtract 1 both sides

`-1=2x+1-y-1`

`-1=2x-y`

Rewrite

`2x-y=-1` ----> equation in standard form