What is the area inscribed by the lines y =3, x = 2, y = x+3 on an xy-coordinate plane?	Select:
A) a) 1 B) b) 2 C) c) 4 D) d) 6 E) e) 10
First, let's graph the lines y = 3 and x = 2   At this point, we need to find the points where the line y = x+3 INTERSECTS the other two lines.   For the vertical line, we know that x = 2, so we'll PLUG x = 2 into the equation y = x+3 to get y = 2+3= 
<gadget id="calculator">2+3</gadget>
<output>5</output>
5   Perfect, when x= 2, y = 5, so one point of intersection is (2,5)   For the horizontal line, we know that y = 3, so we'll PLUG y = 3 into the equation y = x+3 to get 3 = x+3. Solve to get: x = 0   So, when y = 3, x = 0, so one point of intersection is (0,3)   Now add these points to our graph and sketch the line y = x+3   At this point, we can see that we have the following triangle.   The base has length 2 and the height is 2   Area = (1/2)(base)(height)   = (1/2)(2)(2)   = 2   Answer:   b) 2
<result>B</result>
What is the area inscribed by the lines y =3, x = 2, y = x+3 on an xy-coordinate plane?