Given slope and area, how can one find the x & y

Linear, quadratic, module, parametric equations

Given slope and area, how can one find the x & y

Postby Guest » Fri Feb 14, 2020 4:53 am

I am given the area. I also know the slope.
The line is supposed start at 0,0.

I need to know the y coordinate.


I promise this isn't for homework.
I looked up the exact phrase "given slope and area"
and all google comes up with is stuff related to hydrolics.
Guest
 

Re: Given slope and area, how can one find the x & y

Postby Guest » Fri Feb 14, 2020 5:49 am

y = m * sqrt(2*Area/slope)
Guest
 

Re: Given slope and area, how can one find the x & y

Postby HallsofIvy » Mon Feb 17, 2020 10:35 am

I am given the area. I also know the slope.
The line is supposed start at 0,0.

I need to know the y coordinate.


I promise this isn't for homework.

Frankly this doesn't make much sense! You say "I am given the area". The area of what? You say "I also know the slope". The slope of what.

I can guess that you have a line passing through (0, 0), with given slope m, and you want to find the (X, Y) coordinates of another point on that line such that the area of the triangle under the line, the triangle with vertices (0, 0), (X, Y), and (X, 0) is some given number A. The line with slope m, through (0, 0), has equation y= mx. The point with x= X is (X, mX) so the "height" of the triangle is mX, the "base" is X, and the area is "one half height time base", [tex](1/2)(mX)(X)= \frac{m}{2}X^2= A[/tex] to solve that for X, multiply both sides by 2 and divide by m to get [tex]X^2= \frac{2A}{m}[/tex]. Finally, take the square root of both sides- [tex]X= \sqrt{\frac{2A}{m}}[/tex]. And then [tex]Y= mX= m\sqrt{\frac{2A}{m}}= \sqrt{2mA}[/tex].

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