Question

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

Answer 1

Answer:

$y=\frac{3}{2} x+1$

Step-by-step 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 $\frac{rise}{run}$, 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 $\frac{3}{2}$.

Knowing this, the equation of a line parallel to line P would have a slope of $\frac{3}{2}$ as well. Plug this into $y=mx+b$ as m:

$y=\frac{3}{2} x+b$

2) Plug the given point Q into $y=\frac{3}{2} x+b$  to find the y-intercept (b)

$y=\frac{3}{2} x+b$

Plug in point Q (2,4)

$4=\frac{3}{2} (2)+b\\4=\frac{6}{2}+b\\4=3+b$

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:

$y=\frac{3}{2} x+b$

$y=\frac{3}{2} x+1$

I hope this helps!