Question 1155733


given:

lines passing through the point ({{{6}}}, {{{-1}}})
the
product of their x- and y-intercepts is {{{3}}}


if  the product of their x- and y-intercepts is {{{3}}}, one intercept must be {{{1}}} and other must be {{{3}}}

let’s find out which one is {{{1}}} and which one is {{{3}}}


{{{x*y=3}}}

{{{x=3/y}}}

x-intercept at:({{{3/y}}},{{{0}}}) 
y-intercept at:({{{0}}},{{{y}}}) 

then a slope of the line that passes through intercepts is:

{{{m=(y[2]-y[1])/(y[2]-y[1])}}}

{{{m=(y-0)/(0-3/y)}}}

{{{m=y/(-3/y)}}}

{{{m=y^2/-3}}}

plug it in equation of the line:

{{{y=mx+b}}}

{{{y=(y^2/-3)x+b}}}......use given point to calculate {{{b}}}

{{{-1=((-1)^2/-3)*6+b}}}


{{{-1=(1/-1)*2+b}}}

{{{-1=-2+b}}}

{{{2-1=b}}}

{{{b=1}}}-> if {{{b=1}}} then y-intercept is {{{1}}}

=> x-intercept must be {{{3}}}

and {{{m=y^2/-3}}}=>{{{m=1^2/-3}}}=>{{{m=-1/3}}}

your equation will be:


{{{y=-(1/3)x+1}}}


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(6,-1,.12),locate(6,-1,p(6,-1)),
graph( 600, 600, -10, 10, -10, 10, -(1/3)x+1)) }}}