Question

FOUR QUESTIONS!! TYSM IF YOU ARE ANSWERING THIS!! TYSM!!!

Write an equation of the line passing through the given point that is parallel to the given line for the following: (I put a picture bc it’s easier for me and hopefully for you)
THANK YOU SO SO MUCH !!!!

Answer 1

17. Equation :
`y=-x-1`

18: Equation:
`y=(1)/(2)x+8`

19: Equation:
`y=3x-6`

20: Equation:
`y=(1)/(3)x-2`

Explanation:

Using the Slope-Intercept Form for the following questions:

Equation of line:
`y=mx+b` where m is the slope and b is the y-intercept.

17. A(3, -4),
`y=-x+8`

First substitute the slope from the original line
`y=-x+8` (i.e. -1 in this case) into the equation of line.

`y=(-1)x+b` .....(1)

Now, substitute the given point A(3,-4) in the equation (1), we have;

`-4=(-1)\cdot (3)+b`

`-4=-3+b`

`-4+3=b`

∴
`b=-1`

Substitute the value of b=-1 in the Slope-Intercept form i.e,
`y=-x+(-1)`

`y=-x-1`

Therefore, the equation of the line passing through the given point that is parallel to the given line is,
`y=-x-1`.

18. A(-6, 5) ,
`y=(1)/(2)x -7`

First substitute the slope from the original line
`y=(1)/(2)x -7` (i.e.
`(1)/(2)` in this case) into the equation of line.

`y=(1)/(2) x+b` .....(1)

Now, substitute the given point A(-6, 5) in the equation (1), we have;

`5=(1)/(2) \cdot (-6)+b`

`5=-3+b`

`5+3=b`

∴
`b=8`

Substitute the value of b=8 in the Slope-Intercept form i.e,
`y=(1)/(2)x+8`

`y=(1)/(2)x+8`

Therefore, the equation of the line passing through the given point that is parallel to the given line is,
`y=(1)/(2)x+8`.

19. A(2,0),
`y=3x-5`

First substitute the slope from the original line
`y=3x-5` (i.e. 3 in this case) into the equation of line.

`y=3x+b` .....(1)

Now, substitute the given point A(2, 0) in the equation (1), we have;

`0=3\cdot (2)+b`

`0=6+b`

∴
`b=-6`

Substitute the value of b=-6 in the Slope-Intercept form i.e,
`y=3x+(-6)`

`y=3x-6`

Therefore, the equation of the line passing through the given point that is parallel to the given line is,
`y=3x-6`

20. A(3, -1),
`y=(1)/(3)x+10`

First substitute the slope from the original line
`y=(1)/(3)x +10` (i.e.
`(1)/(3)` in this case) into the equation of line.

`y=(1)/(3) x+b` .....(1)

Now, substitute the given point A(3, -1) in the equation (1), we have;

`-1=(1)/(3) \cdot (3)+b`

`-1=1+b`

`-1-1=b`

∴
`b=-2`

Substitute the value of b=-2 in the Slope-Intercept form i.e,
`y=(1)/(3)x+(-2)`

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

Therefore, the equation of the line passing through the given point that is parallel to the given line is,
`y=(1)/(3)x-2`.