Question

Someone help PLEASE

Answer 1

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`