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Write an equation of the line that passes through the point (6,-5) and is parallel to the line whose equation is 2x-3y+11?

1. y + 5 = 3/2 (x-6)2. y = 2/3x + 13. y = -3/2x + 44. y + 5 - 2/3(x-6)

User Anakin
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1 Answer

4 votes

Answer:

Option 4:
(y+5)-(2)/(3)(x-6)=0

Explanation:

It is given that the line parallel to


2x-3y+11=0

hence the slope of this line will the same as the slope of the line that we have to find.

to find the slope of the line, we just to rearrange it in the form y= mx+b


-3y=-2x-11


3y=2x+11


y=(2)/(3)x+(11)/(3)

by comparing it to
y= mx+b. we'll get that m = 2/3

Now we have all that's needed to formulate the equation of our line.

  • the point the line crosses (6,-5)
  • the slope of the line

we'll use the general equation:


(y-y_1)=m(x-x_1), here (x1,y1) = (6,-5)


(y-(-5))=(2)/(3)(x-6)


(y+5)=(2)/(3)(x-6)

we see that this matches option 4.

with a little rearrangement this will be much clearer. We'll move everything on the left hand side of the equation.


(y+5)-(2)/(3)(x-6)=0

User PeYoTlL
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