SOLUTION: 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?

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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?
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

What is the area of the figure bounded by
x+=+-3, x+=+2,
y+%2B+4+=+%28-2%2F5%29%28x+-+2%29, ->y++=+%28-2%2F5%29%28x+-+2%29-4=-%282%2F5%29x-16%2F5+
5y+-+x+-+13+=+0 ->y+=+x%2F5+%2B13%2F5+




when graphed on a coordinate plane,the figure bounded by is a trapezoid

The area is the average of the two base lengths times the altitude:
A=%28%28AB%2BCD%29%2F2%29%2A+h
from graph you see that h=5 (the distance from -3 to 2)
we need to find coordinates of the A,B,C, and D
vertex A is intersection of the lines x+=+-3 and y++=-%282%2F5%29x-16%2F5+
substitute x=-3 in y=-%282%2F5%29x-16%2F5+
y++=+-%282%2F5%29%28-3%29-16%2F5+
y++=+6%2F5-16%2F5+
y++=+-10%2F5+
y++=+-2+
=>A is at (-3,-2)

vertex B is intersection of the lines x+=+-3 and y+=+x%2F5+%2B13%2F5+
substitute x=-3 in y+=+x%2F5+%2B13%2F5+
y++=++-3%2F5+%2B13%2F5+
y++=++10%2F5+
y++=2+
=>B is at (-3,2)


vertex C is intersection of the lines x+=+2 and y++=-%282%2F5%29x-16%2F5+
substitute x=2 in y+=-%282%2F5%29x-16%2F5+
y++=-%282%2F5%292-16%2F5+
y++=+-%284%2F5%29-16%2F5+
y++=+-20%2F5+
y++=+-4+
=>C is at (2,-4)

vertex D is intersection of the lines x+=+2 and y+=+x%2F5+%2B13%2F5+
y+=+2%2F5+%2B13%2F5+
y+=15%2F5+
y+=+3+
=>D is at (2,3)

now find the length of bases AB and CD

AB=sqrt%28%28x%5B2%5D-x%5B1%5D%29%5E2%2B%28y%5B2%5D-y%5B1%5D%29%5E2%29 ...use coordinates of A and B
AB=sqrt%28%28x%5B2%5D-x%5B1%5D%29%5E2%2B%28y%5B2%5D-y%5B1%5D%29%5E2%29
=>A is at (-3,-2)
=>B is at (-3,2)

AB=sqrt%28%28-3-%28-3%29%29%5E2%2B%282-%28-2%29%29%5E2%29
AB=sqrt%28%28-3%2B3%29%5E2%2B%282%2B2%29%5E2%29
AB=sqrt%280%2B4%5E2%29
AB=4

CD=sqrt%28%28x%5B2%5D-x%5B1%5D%29%5E2%2B%28y%5B2%5D-y%5B1%5D%29%5E2%29
=>C is at (2,-4)
=>D is at (2,3)
CD=sqrt%28%282-2%29%5E2%2B%283-%28-4%29%29%5E2%29
CD=sqrt%280%2B%283%2B4%29%5E2%29
CD=sqrt%287%5E2%29
CD=+7

so, h=5 , AB=4, CD=+7, and the area of the figure is:
A=%28%28AB%2BCD%29%2F2%29%2A+h
A=%28%284%2B7%29%2F2%29%2A+5
A=%2811%2F2%29%2A+5
A=55%2F2
A=27.5



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

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?
When plotted, the graphs of: 

a)  and x = - 3 intersect at the point (- 3, 2)

b)  and x = - 3 intersect at the point (- 3, - 2)

c)  and x = 2 intersect at the point (2, 3)

d)  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: