+4609 What is the area of the region bounded by the three lines with equations \(2x+y = 8\), \(2x-5y = 20\) and \(x+y = 10\)?
+128475 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
