Question 1129132
Consider the line {{{x/3 + y/ 8 = k}}}, where {{{k}}} is a positive number


a) Write the coordinates for {{{A}}} and {{{B}}}, the points where the line meets the {{{x}}} and {{{y}}} axes respectively


the points where the line meets the {{{x}}} is where {{{y=0}}}

{{{x/3 + 0/ 8 = k}}}

{{{x  = 3k}}}=>{{{ A}}} ({{{3k}}}, {{{0}}})


the points where the line meets the {{{y}}} is where {{{x=0}}}

{{{0/3 + y/ 8 = k}}}

{{{y/ 8 = k}}}

{{{y = 8k}}}=>{{{B}}}({{{0}}}, {{{8k}}})



b) The area of triangle {{{AOB}}} where{{{ O}}} is the origin, is {{{144}}} units squared. Find the value for{{{ k }}}


The area of triangle {{{AOB}}}: 


{{{area=(base*h)/2}}} where


base is distance from origin to {{{x}}}  the point {{{A}}}
{{{OA=3k}}}

height is distance from origin to {{{y}}}  the point {{{B}}}

{{{OB=8k}}}


so, we have:

{{{144=(3k*8k)/2}}}

{{{144=12k^2}}}

{{{144/12=k^2}}}

{{{12=k^2}}}

{{{k=sqrt(12)}}}

{{{k=sqrt(4*3)}}}

{{{k=2sqrt(3)}}}


then,

{{{OA=3k}}}=>{{{OA=3*2sqrt(3)}}}=>{{{OA=6sqrt(3)}}}

{{{OB=8k}}}=>{{{OB=8*2sqrt(3)}}}>=>{{{OB=16sqrt(3)}}}



check the area:

{{{area=(base*h)/2}}} 
 
{{{area=(6sqrt(3)*16sqrt(3))/2}}}

{{{area=6sqrt(3)*8sqrt(3)}}}

{{{area=6*8(sqrt(3))^2}}}

{{{area=6*8*3}}}

{{{area=144}}}