Question 313871
Graph the triangle enclosed by the Y-axis and the 
graphs of y - 3x = -2 and y + 3x = 2 
Find the area of the triangle.
<pre><b>
Draw the graphs of the lines:

{{{system(y-3x=-2,y+3x=2)}}}

First put them in slope intercept form:

{{{system(y=3x-2,y=-3x+2)}}}

Their y-intercepts are (0,-2) and (0,2) respectively,

So these are two of the vertices of the triangle we wish

to find the area of. 

We solve the system of equations to find the coordinates
wheteh they intersect, which be the third vertex of the
triangle we wish to find the area of.

{{{system(y=3x-2,y=-3x+2)}}}

Set the values of y equal:

{{{3x-2=-3x+2}}}
{{{6x=4}}}
{{{x=2/3}}}

Substituting in 

{{{y=3x-2}}}
{{{y=3(2/3)-2}}}
{{{y=cross(3)(2/cross(3))-2
{{{y=2-2}}}
{{{y=0}}}

So the third point of the triangle is (0,{{{2/3}}})

{{{system(y=3x-2,y=-3x+2)}}}

Draw the graphs of the lines

{{{drawing(400,400,-3,3,-3,3,
locate(0.1,2,"(0,2)"), locate(0,-2,"(0,-2)"), locate(2/3,.3,"(2/3,0)"),
graph(400,400,-3,3,-3,3,3x-2), graph(400,400,-3,3,-3,3,-3x+2) )}}} 

So we wish to find the area of this triangle:

{{{drawing(400,400,-3,3,-3,3, graph(400,400,-3,3,-3,3),
locate(0.1,2,"(0,2)"), locate(0,-2,"(0,-2)"), locate(2/3,.3,"(2/3,0)"),
line(0.04,-2,0.04,2), line(0.04,-2,2/3,0), line(0.04,2,2/3,0)
 )}}} 

Take as a base b the line segement from (0,-2) to (0,2) which is
4 units long.  Take as an altitude h the line segment from (0,0) to
({{{2/3}}},0) which is {{{2/3}}} units long.

Area of a triangle is  

A = {{{1/2}}}bh 

A = {{{1/2}}}×(4)×({{{2/3}}})

A = {{{1/cross(2)}}}×(4)×({{{cross(2)/3}}})

A = {{{4/3}}}

Edwin</pre>