Question

What is the equation of a line in slope-intercept form that is perpendicular to the equation
y = 4x + 3 and goes through the point (-4, 5) ?

Answer 1

Answer:

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Step-by-step explanation:

So, we must find the equation of a line perpendicular to the given line and point such that,

we take the given slope and take the reciprocal of it such that m is now -1/4 and we pull out the point slope formula such that

y - y1 = m(x - x1)

y - 5 = -1/4(x - (-16))

y - 5 = -1/4(x + 16)..............Distribute the -1/4 inside such that

y - 5 = -1/4x - 4............then add 5 such that

y = (-1/4)x + 1

I hope this helped!

Answer 2

Answer:

y=-1/4x+4

Step-by-step explanation:

Find the equation of the line that is

perpendicular to y = 4x+3

and passes though the point (-4,5)

The slope of y=4x+3is: 4

The negative reciprocal of that slope is:

m = −1/4 = -1/4

So the perpendicular line will have a slope of -1/4:

y − y1 = (-1/4)(x − x1)

And now put in the point (-4,5):

y − 5= (-1/4)(x + 4)

And that answer is OK, but let's also put it in "y=mx+b" form:

y − 5= (-1/4)(x + 4)

y=-1/4x+4