Answer:
[tex]y=\frac{3}{2} x+1[/tex]
Step-by-step explanation:
What we need to know:
- Linear equations are typically organized in slope-intercept form: [tex]y=mx+b[/tex] where m is the slope and b is the y-intercept (the value of y when the line crosses the y-axis)
- 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 [tex]\frac{rise}{run}[/tex], 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 [tex]\frac{3}{2}[/tex].
Knowing this, the equation of a line parallel to line P would have a slope of [tex]\frac{3}{2}[/tex] as well. Plug this into [tex]y=mx+b[/tex] as m:
[tex]y=\frac{3}{2} x+b[/tex]
2) Plug the given point Q into [tex]y=\frac{3}{2} x+b[/tex] to find the y-intercept (b)
[tex]y=\frac{3}{2} x+b[/tex]
Plug in point Q (2,4)
[tex]4=\frac{3}{2} (2)+b\\4=\frac{6}{2}+b\\4=3+b[/tex]
Subtract 3 from both sides
[tex]4-3=3+b-3\\1=b[/tex]
Therefore, the y-intercept of this line (b) is 1. Plug this back into our original equation:
[tex]y=\frac{3}{2} x+b[/tex]
[tex]y=\frac{3}{2} x+1[/tex]
I hope this helps!
