Question 1152213

What is the area of the figure bounded by x = -3,  x = 2,  y + 4 = (-2/5)(x - 2),  and 5y - x - 13 = 0 when graphed on a coordinate plane?
<pre>When plotted, the graphs of: 
a) {{{matrix(1,7, 5y - x - 13, "=", 0, or, y, "=", (1/5)x + 13/5)}}} and x = - 3 intersect at the point (- 3, 2)
b) {{{matrix(1,7, y + 4, "=", (- 2/5)(x - 2), or, y, "=", (- 2/5)x - 16/5)}}} and x = - 3 intersect at the point (- 3, - 2)
c) {{{matrix(1,7, 5y - x - 13, "=", 0, or, y, "=", (1/5)x + 13/5)}}} and x = 2 intersect at the point (2, 3)
d) {{{matrix(1,7, y + 4, "=", (- 2/5)(x - 2), or, y, "=", (- 2/5)x - 16/5)}}} and x = 2 intersect at the point (2, - 4)
These four points form a trapezoid, with a height of 5 (2 - - 3), as seen on the x-axis
With the trapezoid's shorter base being 4 [from (- 3, 2) to (- 3, - 2)] units, and longer base being 7 [from (2, 3) to (2, - 4)], 
we then get the trapezoid's area or the area bounded by the four line graphs as: {{{highlight_green(matrix(1,5, (1/2)(4 + 7)(5), "=", 27.5, square, units))}}}