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

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