# Calculate the area of ​​a figure bounded by lines x-y+3=0 x+y-1=0 y=0

Calculate the area of ​​a figure bounded by lines
x-y+3=0 x+y-1=0 y=0
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Clarence Mills
Three lines form a triangle.
The line x-y+3=0 passes through the points (-3; 0) and (0; 3)
The line x+y-1=0 passes through the points (0;1) and (1;0)
y=0 - the equation of the x-axis.
Direct
x-y+3=0 and
x+y-1=0
intersect at x=1 y=2
Triangle isosceles
Base from point -3 to point 1
The height passes through the intersection point x=1 y=2 and is equal to the ordinate of this point
S=1/2 4 2=4 sq. units
Second way
$S={\int }_{-3}^{-1}\left(x+3\right)dx+{\int }_{-1}^{1}\left(-x+1\right)dx=\left(\frac{{x}^{2}}{2}+3x{\right)}_{-3}^{-1}+\left(\frac{-{x}^{2}}{2}+x{\right)}_{-1}^{1}=$
$=\frac{1}{2}-3-\left(\frac{9}{2}-9\right)+\left(-\frac{1}{2}+1\right)-\left(-\frac{1}{2}-1\right)=4$