Question 1152213

What is the area of the figure bounded by 
{{{x = -3}}}, {{{x = 2}}}, 
{{{y + 4 = (-2/5)(x - 2)}}}, ->{{{y  = (-2/5)(x - 2)-4=-(2/5)x-16/5 }}}
{{{5y - x - 13 = 0}}} ->{{{y = x/5 +13/5 }}}



{{{drawing ( 600, 600, -10, 10, -10, 10,
line(-3,10,-3,-10),line(2,10,2,-10),locate(-3.5,-2,A),locate(-3.5,2.5,B),locate(2.3,-4.3,C),locate(2.3,4,D),circle(-3,-2,.13),locate(-3,-2,p(-3,-2)),
circle(-3,2,.13),locate(-3,2,p(-3,2)),
circle(2,-4,.13),locate(2,-4,p(2,-4)),
circle(2,3,.13),locate(2,3,p(2,3)),
graph( 600, 600, -10, 10, -10, 10, (-2/5)(x - 2)-4, x/5 +13/5 )) }}}


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=((AB+CD)/2)* 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  =-(2/5)x-16/5 }}}

substitute {{{x=-3}}} in {{{y=-(2/5)x-16/5 }}}

{{{y  = -(2/5)(-3)-16/5 }}}

{{{y  = 6/5-16/5 }}}

{{{y  = -10/5 }}}

{{{y  = -2 }}}

=>{{{A}}} is at ({{{-3}}},{{{-2}}})


vertex {{{B}}} is intersection of the lines {{{x = -3}}} and {{{y = x/5 +13/5 }}}

substitute {{{x=-3}}} in {{{y = x/5 +13/5 }}}

{{{y  =  -3/5 +13/5 }}}

{{{y  =  10/5 }}}

{{{y  =2 }}}

=>{{{B}}} is at ({{{-3}}},{{{2}}})



vertex {{{C}}} is intersection of the lines {{{x = 2}}} and {{{y  =-(2/5)x-16/5 }}}

substitute {{{x=2}}} in {{{y =-(2/5)x-16/5 }}}

{{{y  =-(2/5)2-16/5 }}}

{{{y  = -(4/5)-16/5 }}}

{{{y  = -20/5 }}}

{{{y  = -4 }}}

=>{{{C}}} is at ({{{2}}},{{{-4}}})


vertex {{{D}}} is intersection of the lines {{{x = 2}}} and {{{y = x/5 +13/5 }}}

{{{y = 2/5 +13/5 }}}

{{{y =15/5 }}}

{{{y = 3 }}}

=>{{{D}}} is at ({{{2}}},{{{3}}})


now find the length of bases {{{AB}}} and {{{CD}}}


{{{AB=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}} ...use coordinates of {{{A}}} and {{{B}}}

{{{AB=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

=>{{{A}}} is at ({{{-3}}},{{{-2}}})
=>{{{B}}} is at ({{{-3}}},{{{2}}})


{{{AB=sqrt((-3-(-3))^2+(2-(-2))^2)}}}

{{{AB=sqrt((-3+3)^2+(2+2)^2)}}}

{{{AB=sqrt(0+4^2)}}}

{{{AB=4}}}


{{{CD=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

=>{{{C}}} is at ({{{2}}},{{{-4}}})
=>{{{D}}} is at ({{{2}}},{{{3}}})

{{{CD=sqrt((2-2)^2+(3-(-4))^2)}}}

{{{CD=sqrt(0+(3+4)^2)}}}

{{{CD=sqrt(7^2)}}}

{{{CD= 7}}}


so, {{{h=5}}} , {{{AB=4}}}, {{{CD= 7}}}, and the area of the figure is:

{{{A=((AB+CD)/2)* h}}}

{{{A=((4+7)/2)* 5}}}

{{{A=(11/2)* 5}}}

{{{A=55/2}}}

{{{A=27.5}}}