# Find the area of a triangle.

• April 2nd 2012, 03:09 PM
joshuaa
Find the area of a triangle.
Find the area of a triangle enclosed by axes and tangent to y = 1/x

(Crying)
• April 2nd 2012, 03:13 PM
MathVideos
Re: Find the area of a triangle.
Where do you run into your problem? Finding the tangent line? Also, there's an infinite amount of triangles that could be made. Do they specify anything else?
• April 2nd 2012, 03:35 PM
joshuaa
Re: Find the area of a triangle.

Attachment 23501
• April 2nd 2012, 07:10 PM
Prove It
Re: Find the area of a triangle.
Quote:

Originally Posted by joshuaa

Attachment 23501

Well, let's see. If the function is \displaystyle \begin{align*} y = \frac{1}{x} \end{align*}, the derivative is \displaystyle \begin{align*} \frac{dy}{dx} = -\frac{1}{x^2} \end{align*}. This tells us the gradient of the tangent line at any point x.

So the tangent line is of the form \displaystyle \begin{align*} y = \left(-\frac{1}{x^2}\right)x + c = -\frac{1}{x} + c \end{align*}.

c is the y intercept. To find the x intercept, we'll let y = 0, and we find that

\displaystyle \begin{align*} 0 &= -\frac{1}{x} + c \\ \frac{1}{x} &= c \\ x &= \frac{1}{c} \end{align*}.

So the height of the triangle is c, the length is \displaystyle \begin{align*} \frac{1}{c} \end{align*}, which means the area is

\displaystyle \begin{align*} A &= \frac{1}{2}(c)\left(\frac{1}{c}\right) \\ &= \frac{1}{2} \end{align*}

So the area is \displaystyle \begin{align*} \frac{1}{2}\,\textrm{unit}^2 \end{align*}
• April 3rd 2012, 12:01 AM
biffboy
Re: Find the area of a triangle.
Don't think that is correct. Example consider tangent at (1,1) Gradient is -1 Equation of tangent is y=-x+2 Intercepts are both 2 Area of triangle is 2
In general consider tangent at (k,m) Gradient of tangent is -1/k^2 Get the equation of the tangent and the intercepts. Knowing also that km=1 wil get area of triangle=2
• April 3rd 2012, 12:16 AM
alexmahone
Re: Find the area of a triangle.
Quote:

Originally Posted by Prove It
So the tangent line is of the form \displaystyle \begin{align*} y = \left(-\frac{1}{x^2}\right)x + c = -\frac{1}{x} + c \end{align*}.

This isn't the equation of a line. (See post #5.)
• April 3rd 2012, 03:11 AM
joshuaa
Re: Find the area of a triangle.
Thank you very much. That was helpful :)