Answer :
Answer:
[tex]y+3=\frac{5}{3}(x-9)[/tex] or [tex]y=\frac{5}{3}x-18[/tex]
Step-by-step explanation:
step 1
Find the slope of the line parallel to the given line
we know that
If two lines are parallel, then their slopes are the same
we have
[tex]y=\frac{5}{3}x-4[/tex]
therefore
The slope m of the line parallel to the given line is
[tex]m=\frac{5}{3}[/tex]
step 2
Find the equation of the line in point slope form
[tex]y-y1=m(x-x1)[/tex]
we have
[tex]m=\frac{5}{3}[/tex]
[tex]point\ (9,-3)[/tex]
substitute
[tex]y+3=\frac{5}{3}(x-9)[/tex] ----> equation of the line in point slope form
step 3
Find the equation of the line in slope intercept form
[tex]y=mx+b[/tex]
we have
[tex]y+3=\frac{5}{3}(x-9)[/tex]
Isolate the variable y
[tex]y+3=\frac{5}{3}x-15[/tex]
[tex]y=\frac{5}{3}x-15-3[/tex]
[tex]y=\frac{5}{3}x-18[/tex]
Other Questions
If the ka of a monoprotic weak acid is 1.2 × 10-6, what is the ph of a 0.40 m solution of this acid?
If the ka of a monoprotic weak acid is 1.2 × 10-6, what is the ph of a 0.40 m solution of this acid?