Question

What is an equation of the line that is perpendicular to -4/5x+3 and passes through the point(4, 12)

Answer 1

Now, for the equation that we're trying to find, all that they give us is their slope, and a point that it passes through. How do we get the slope? Perpendicular line's slopes are negative reciprocals of each other. This means that you flip the coefficient of the x around, and multiply it by negative 1. So, we see that the given slope of the other line is -4/5. That means, we flip it around (-5/4), them multiply it by -1, which means the slope of our new line is 5/4. Now, we have the x, but what do we do with the point? We plug it into the point-slope formula

y-y1=m(x-x1)      , where m = slope, and (x1,y1)

Our slope (m) is 5/4, our x1 is 4, and our y2 is 12 (these are the points given to us in the original problem, (4,12).)

Now, we plug it in, then get y by itself

y-y1=m(x-x1)
y-(12)=(5/4)(x-4)
y-12=5/4(x-4)
y=5/4(x-4)+12
y=5/4x-5+12
y=5/4x+7

So, our new equation is

y=5/4x+7