How to find the area of the triangle formed by the lines $y=ax$ , $x+y-a=0$ and the $y$ axis? I found out the intersection points $A( \frac{a}{1+a}$ , $\frac{a^2}{1+a} )$ and $B$ as $(0,a)$. Now, I don't know what to do next. Please explain in easy steps. Thank you!
 A: First, you should draw all these lines on the $Oxy$ plane. Then, it is very easy to see the area of this triangle, whose the 1st vertex is your A, 2nd is $D(a;0)$, and 3rd is $O(0;0)$.
The area is $\frac{1}{2}*base*height = \frac{1}{2}*a*\frac{a}{1+a} = \frac{a^2}{2(1+a)}$
A: By solving the equations of lines: $y=ax$, $x+y-a=0$ & $x=0$, we can easily determine the vertices of the given triangle are: $A\left(\frac{a}{1+a}, \frac{a^2}{1+a}\right)$, $B(0, a)$ & $C(0, 0)$ 
Then the area($\Delta$) of a triangle, having its vertices $(x_{1}, y_{1})$, $(x_{2}, y_{2})$ & $(x_{3}, y_{3})$, is given by the generalized formula of area as follows $$\bbox[4pt, border: 1px solid blue; ]{\Delta=\left|\frac{1}{2} \left(x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right)\right|}$$ Setting the corresponding values in the above formula, we get area of triangle
$$\Delta=\left|\frac{1}{2} \left(\frac{a}{1+a}(a-0)+(0)\left(0-\frac{a^2}{1+a}\right)+(0)\left(\frac{a^2}{1+a}-a\right)\right)\right|$$ $$=\frac{a^2}{2(1+a)}$$
