[tex]\huge\boxed{y+2=\frac{3}{5}(x-15)}[/tex]
We will use point-slope form, where [tex]m[/tex] is the slope and [tex](x_1,y_1)[/tex] is a known point on the line.
[tex]y-y_1=m(x-x_1)[/tex]
First, find the slope. Since the two lines will be parallel, they'll share the same slope of [tex]\frac{3}{5}[/tex].
[tex]y-y_1=\frac{3}{5}(x-x_1)[/tex]
Substitute in the known point.
[tex]y-(-2)=\frac{3}{5}(x-15)[/tex]
Simplify the negative subtraction.
[tex]\large\boxed{y+2=\frac{3}{5}(x-15)}[/tex]
Note: If you need the answer in a different form, please let me know below.
Answer:
y = [tex]\frac{3}{5}[/tex] x - 11
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
y = [tex]\frac{3}{5}[/tex] x - 3 ← is in slope- intercept form
with slope m = [tex]\frac{3}{5}[/tex]
Parallel lines have equal slopes, thus
y = [tex]\frac{3}{5}[/tex] x + c ← is the partial equation of the parallel line
To find c substitute (15, - 2) into the partial equation
- 2 = 9 + c ⇒ c = - 2 - 9 = - 11
y = [tex]\frac{3}{5}[/tex] x - 11 ← equation of parallel line
