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What is the area of the region bounded by the three lines with equations \(2x+y = 8\)\(2x-5y = 20\) and \(x+y = 10\)?

 Dec 29, 2017
 #1
avatar+98107 
+2

Look  at the graph, here :

 

https://www.desmos.com/calculator/sdppfsq0xq

 

The three intersection points  occur at  (5,-2), (10, 0)  and (-2, 12)

 

And these form a triangle......

 

Let us  consider  the base to be the segment joining  (10,0)  and (-2,12)

 

And the length  of this base  =  sqrt [ (-2 -10)^2  + 12^2]  =   sqrt (288)  =  12sqrt(2)

 

And the altitude of this triangle can be drawn from (5, -2)  perpendicular to the line x + y  - 10 = 0

 

And this distance can be calculated as follows :

 

Altitude  =    abs   [  ax + by  -  10 ]  /  sqrt ( a^2  + b^2 ) 

Where (a, b)  = (1, 1)    and  (x, y)  = (5, -2)

 

So we have

 

Altitude  =  abs  [ 5  +  (-2)   -  10] / sqrt [  1^2 + 1^2 ]  =

 

abs  [ -7]  /  sqrt [ 2]   =      7 / sqrt (2)

 

So.....this area  =  (1/2)base * height  =

 

(1/2)(12*sqrt(2) ) *  7 / sqrt (2)   =   42 units^2

 

 

 

cool cool cool

 Dec 29, 2017
 #2
avatar+3994 
+1

Oh, I get it now! Thanks again, CPhill! smileycool

 Dec 29, 2017

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