find the area of the triangle whose sides are along the lines x=3,y=0a

1.	find the area of the triangle whose sides are along the lines x=3,y=0and 4x+5y=20.
Answer» A is point of intersection of line x = 2 and 4x + 5y = 20{tex}\\Rightarrow \\quad 4 \\times 2 + 5 y{/tex}\xa0= 20{tex}\\Rightarrow \\quad y = \\frac { 12 } { 5 }{/tex}{tex}\\therefore{/tex}\xa0Coordinates of A are\xa0{tex}\\left( 2 , \\frac { 12 } { 5 } \\right){/tex}B is the point of intersection of x = 2 and y = 0{tex}\\therefore{/tex}\xa0Coordinates of B are (2, 0).C is point of intersection\xa0y = 0 and 4x + 5y = 20{tex}\\Rightarrow \\quad 4 x + 5 \\times 0{/tex}\xa0= 20{tex}\\Rightarrow{/tex}\xa0x = 5{tex}\\Rightarrow{/tex}\xa0Coordinates of C are (5, 0)Area\xa0{tex}\\triangle {/tex}ABC = {tex}\\frac { 1 } { 2 } \\left\| x _ { 1 } \\left( y _ { 2 } - y _ { 3 } \\right) + x _ { 2 } \\left( y _ { 3 } - y _ { 1 } \\right) + x _ { 3 } \\left( y _ { 1 } - y _ { 2 } \\right)\\right.{/tex}=\xa0{tex}\\frac { 1 } { 2 } \\left\| 2 ( 0 - 0 ) + 2 \\left( 0 - \\frac { 12 } { 5 } \\right) + 5 \\left( \\frac { 12 } { 5 } - 0 \\right) \\right\|{/tex}=\xa0{tex}\\frac { 1 } { 2 } \\left\| \\frac { - 24 } { 5 } + 12 \\right\|{/tex}=\xa0{tex}\\frac { 1 } { 2 } \\times \\frac { 36 } { 5 }{/tex}=\xa0{tex}\\frac { 18 } { 5 }{/tex}\xa0sq. units= 3.6 sq. units

find the area of the triangle whose sides are along the lines x=3,y=0and 4x+5y=20.