Question 1151945
a triangle has an area of {{{10}}} square units and enclosed by the {{{x}}}-axis, {{{y}}}-axis and the line {{{2y+ax-10=0}}}

Find the values of {{{a}}}

given:

a triangle has an area of {{{10}}}

{{{x=0 }}} is y-axis, and {{{y=0}}} is x-axis.


{{{2y+ax-10=0}}}=> if {{{x=0}}}},

{{{2y=10}}}

{{{y=5}}}=>y-intercept is at ({{{0}}},{{{5}}})=> one leg of triangle (height) is {{{5}}}


{{{ax=10-2y}}}=> if {{{y=0}}}}

{{{ax=10-2*0}}}

{{{ax=10}}}


=>{{{x=10/a}}}=> the other leg of triangle, x- intercept is at ({{{10/a}}},{{{0}}})


since a triangle has an area of {{{10}}}, we have

{{{10=(1/2)*5(10/a)}}}

{{{10=50/2a}}}

{{{10*2a=50}}}

{{{2a=5}}}

{{{a=5/2}}}

and, your equation is:

{{{2y+(5/2)x-10=0}}}

then x- intercept is at ({{{10/a}}},{{{0}}})=({{{4}}},{{{0}}})

check the area:

{{{10=(1/2)*5(10/(5/2))}}}

{{{10=(1/2)*5(20/5)}}}

{{{10=(1/2)*20}}}

{{{10=10}}}


{{{drawing ( 600, 600, -10, 10, -10, 10, 
circle(0,5,.12),circle(4,0,.12),
locate(0.3,5,p(0,5)),locate(4,-0.5,p(4,0)),locate(1,1,leg=4),locate(-2,2.5,leg=5),
graph( 600, 600, -10, 10, -10, 10, -(5/4)x+5)) }}}