# A right triangle in the first quadrant is bounded by lines y=0, y=x, and y= -x+5. Find its area?

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y= 0 y= x y= -x+5

First we will find the coordinated of the vertices's.

To do that we will determine the points of intersections:

y= 0 and y= x

==> x= 0 ==> y= 0

==>A (0, 0) is one of the vertices's>

y= 0 and y= -x + 5

==> -x + 5 = 0

==> x = 5

==> B(5, 0) is one of the vertices's>

y= x and y= -x + 5

==> x= -x + 5

==> 2x= 5

==> x = 5/2

==> y= 5/2

==> C(5/2 , 5/2) is the third vertices's

==> A(0,0) B(5,0) and C( 5/2, 5/2) is aright angle triangle:

==> AB = sqrt ( 5-0)^2 + 062 = 5

==> AC = sqrt( (5/2)^2 + (5/2)62 = sqrt(50/4) = (5/2)sqrt2

==> BC = sqrt((5-5/2)^2 + (5/2)^2 = (5/2)sqrt2

The AB is the largest then AB is the hypotenuse:

==> A = (1/2) * AC* BC = (1/2)*(5/2)sqrt2 * (5/2)sqrt2

==> A = 25/8 * 2

**==> A = 25/4**

The triangle is made by the lines y = 0, y = x and y = -x +5.

Now we find the point of intersection of the lines y = x and y = -x + 5. We have x = -x +5

=> 2x = 5

=> x = 5/2

=> y = 5/2

Now the lines y=0 and y = x intersect at the origin.

The distance of the point (5/2 , 5/2) from the origin is (5/2)*sqrt 2.

The point where y = 0 and y = -x + 5 intersect is (5, 0)

The distance between (5/2, 5/2 ) and (5, 0) is (5/2) sqrt 2.

Therefore the area of the triangle is (1/2)*base*height

= (1/2)* (5/2) * sqrt 2 * (5/ 2) *sqrt 2

= 25 /4

= 6.25

**The area of the traingle is 6.25**