Question

What is the equation of the line that passes through point Q and is parallel to line P?​

Answer 1

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

Explanation:

What we need to know:

1. Linear equations are typically organized in slope-intercept form:
   `y=mx+b` where m is the slope and b is the y-intercept (the value of y when the line crosses the y-axis)
2. Parallel lines always have the same slope (m) and different y-intercepts (b)

1) Find the slope of line P

The slope of a line is equal to the
`(rise)/(run)`, or the number of units the line moves up over the number of units the line moves to the right.

We can see that for line P, for every 2 units it moves to the right, it moves up 3 units.

Therefore, the slope of line P is
`(3)/(2)`.

Knowing this, the equation of a line parallel to line P would have a slope of
`(3)/(2)` as well. Plug this into
`y=mx+b` as m:

`y=(3)/(2) x+b`

2) Plug the given point Q into
`y=(3)/(2) x+b` to find the y-intercept (b)

`y=(3)/(2) x+b`

Plug in point Q (2,4)

`4=(3)/(2) (2)+b\\4=(6)/(2)+b\\4=3+b`

Subtract 3 from both sides

`4-3=3+b-3\\1=b`

Therefore, the y-intercept of this line (b) is 1. Plug this back into our original equation:

`y=(3)/(2) x+b`

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

I hope this helps!