XXQ68 Find the area in square units of the triangle formed by 2x 3y 5 y x and X-Axis

the points of intersection for lines:
2x+3y=5 and x=y => 5x=5 => x=1 and y=1 =>(1,1)
x=y and y=0(x-axis) => x=0 and y=0 =>(0,0)
2x+3y=5 and y=0(x-axis) => 2x=5 => x=5/2 and y=0 =>(5/2,0)

area of triangle with vertices (x1,y1),(x2,y2),(x3,y3) is
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|/2
=|1(0-0)+0(0-1)+5/2(1-0)|/2
=|1+5/2|/2
=7/4 sq units...
12 years agoHelpfull: Yes(19) No(10)
rewrite the eq.as x/(5/2) + y/(5/3) = 1
intersection at X and Y axes are 5/2 and 5/3.
so, base of triangle= 5/2
and intersction points are (1,1)
therefore, height is 1.
area = 1/2*base*height
=1/2*5/2*1 = 5/4 sq units
12 years agoHelpfull: Yes(11) No(1)
as per sainath's approch.,.,
the co-ordinates are (1,1),(0,0)(5/2,0),..,

now.,. area of triangle =1/2*1*5/2
=== > 5/4..,
12 years agoHelpfull: Yes(7) No(1)
ans - 25/12
as it is on x-axis so base=5/2 and height=5/3
12 years agoHelpfull: Yes(1) No(13)
Dear Saitath can u explain how we can get 0(0-1)=1????
12 years agoHelpfull: Yes(1) No(2)
plz xplain the right calculation vivek..
12 years agoHelpfull: Yes(0) No(1)
m getting the same ans as sainath has got...thats y i posted it...
@vivek tiwari can u xplain d calculation..
12 years agoHelpfull: Yes(0) No(1)

XXQ68) Find the area (in square units) of the triangle formed by 2x+3y=5, y=x and X-Axis.