Question 735628
{{{m(x)= -5x + 123/7}}} and {{{x+5y= -12 }}}

firs write them in slope-intercept form: {{{y=mx+b}}} where {{{m=slope}}} and {{{b=y-intercept}}}

{{{m(x)= -5x + 123/7}}}..already in  slope-intercept form: {{{m=-5}}} and {{{b= 123/7}}}       


{{{x+5y= -12 }}}...solve for {{{y}}}

{{{5y= -x-12 }}}  

{{{y= (-1/5)x-12/5 }}}  ....{{{m=-1/5}}} and {{{b= 12/5}}}     

take a look at slopes:    {{{m=-5}}} and  {{{m=-1/5}}}

lines that are parallel have {{{same}}} {{{slope}}}: since  {{{-5<>-1/5}}}, these line are not parallel

to be perpendicular, lines have to have slopes negative reciprocals of each other
as you can see, {{{5}}} and {{{1/5}}} are reciprocals, but you have {{{-5}}} and {{{-1/5}}}; so, since both negative, one is {{{not}}} negative reciprocal of other one

therefore,these line are not perpendicular ether 

so, these lines are neither parallel nor perpendicular,  they must be intersecting in one point

let's check it on a graph:


               {{{ graph( 600, 600, -10, 10, -10, 10, -5x + 123/7, (-1/5)x-12/5) }}}