189k views
2 votes
Someone help PLEASE

Someone help PLEASE-example-1

1 Answer

2 votes

Answer:

Explanation:

For part a, you are given a slope of
(-5)/(6) x. To find a perpendicular line, you must find the negative reciprocal of that slope. Note that the "+3" in the original equation provided does not mean anything.

To find the negative reciprocal, simply swap the -5 and 6 from each other's positions as a numerator and denominator and multiply by a negative.


-((-6)/(5) )


(6)/(5)

Now, you can use the point-slope formula to plug in a point and slope. In this case you are given a point and the slope you found earlier.


y-(y-coord) = m(x-(x-coord))


y-(-13) = (6)/(5) (x-0)


y+13 = (6)/(5)x\\\\ y = (6)/(5)x -13

For part b, I would first find the isolate y and put the equation in the form y = mx+b. Then, I would repeat the steps shown above.


-4x + 5y = 15\\5y = 15+4x\\y = (4)/(5) +3

Negative Reciprocal.


-((5)/(4))\\\\-(5)/(4)

Point Slope.


y-(11) = -(5)/(4) (x-0)\\y-11 = -(5)/(4)x\\y = -(5)/(4)x + 11

User Dzinx
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.