Question 935864
If a parallelogram has two consecutive sides {{{congruent}}}, it is a rhombus. Also oppo 

if the {{{3}}} other sides lie on {{{3x +2y + 12 =0}}},{{{ 4x -3y -12=0}}} and {{{6x +4y -11=0}}}, then fourth side lie on {{{ax+by+c=0}}} 

write these lines in slope-intercept form
{{{3x +2y + 12 =0}}}=> {{{ y =-(3/2)x-6}}}=> a slope is {{{-(3/2)}}}

{{{ 4x -3y -12=0}}} =>{{{ y =(4/3)x-4}}}=> a slope is {{{(4/3)}}}

these lines are not parallel, they have different slopes; it means, third given line must be parallel to one of these lines

find a slope of third line:{{{6x +4y -11=0}}}=>{{{y=-(6/4)x+11/4 }}}=>{{{y=-(3/2)x+11/4 }}}=> a slope is {{{-(3/2)}}}

so, lines {{{ y =-(3/2)x-6}}} and {{{y=-(3/2)x+11/4 }}} are parallel

now, we need to find the distance between these parallel lines using distance formula: find one point on each line
if {{{x=0}}} then {{{ y =-6}}} :({{{0}}},{{{-6}}})
if {{{x=0}}} then {{{y=11/4 }}}:({{{0}}},{{{11/4=2.75}}})

{{{d=sqrt((0-0)^2+(2.75-(-6))^2)}}}

{{{d=sqrt((2.75+6)^2)}}}

{{{d=8.75}}}



then we need to find a fourth line {{{y=mx+b}}} which is parallel to {{{ y =(4/3)x-4}}} and will have a slope {{{(4/3)}}} and {{{d=8.75}}}

{{{y=(4/3)x+b}}}

find one point on {{{ y =(4/3)x-4}}}

if {{{x=0}}} then {{{ y =-4}}}:({{{0}}},{{{-4}}})

{{{8.75=sqrt((y-(-4))^2)}}}

{{{8.75=y+4}}}

{{{8.75-4=y}}}

{{{4.75=y}}}...point on a line we are looking for is :({{{0}}},{{{4.75}}}) which is y-intercept or {{{b}}}

so, our line is {{{highlight(y=(4/3)x+4.75)}}}

{{{ graph( 600, 600, -10, 10, -10, 10,-3x/2-6, 4x/3-4,-6x/4+2.75,(4/3)x+4.75) }}}