Find the area of the region between two lines and $ x = 0$ I need to know the area between $y = 2x+4$ and $y = 4x$. Between each other before they intercept at $(2, 8)$. That's against the $y$-axis.
this will help  
 A: if you take the $y$-intercept of $4$ as the base, then the height is the $x$-coordinate of $2$ of the point $(2,8).$  so the area of the yellow triangle is $$\frac 12 \times 4 \times 2 = 4$$
A: There are 2 ways you can get the area.
From the point of intersection, p$(2,8)$, drop a perpendicular on the Y-axis, giving 8 units as the result.
Base is 4 units. The length of the side, lying on the y axis. 
Then, the area is given by
$$A=\frac{1}{2}\times 4 \times 8= 16$$
You can also use the Hero's formula. Because, you know all the coordinates of the triangle, and using them you can get the sides. Calculate the semi-perimeter
$$s=\frac{a+b+c}{2}$$
and use below, to calculate the area
$$Area=\sqrt{s(s-a)(s-b)(s-c)}$$
A: You have a rectangle with vertices $(0,0),(2,0),(8,0)$,and$(8,2)$. The rectangle, which has an area of $8*2=16$, is divided into three triangles, with the middle one being the area you're looking for. The 'top' triangle has a base of 4 and height of 2, so it's $0.5*4*2=4$. The 'bottom' triangle has a base of 2 and a height of 8, so its area is $0.5*8*2=8$. So the area of the triangle you're looking for is $16-8-4=4$.
