Question 1170509


Lines {{{l}}} and {{{l}}}′ with equations

{{{y=(11k+17)x+1}}}....slope: {{{(11k+17)}}}


{{{x=(61k-23)y-2}}}.........solve for {{{y}}}

{{{y=x/(61k-23)+2}}}

{{{y=(1/(61k-23))x+2}}}....slope: {{{1/(61k-23)}}}


since lines are {{{perpendicular}}}, slopes are {{{negative}}}{{{ reciprocal}}} to each other


{{{(11k+17) =-1/(1/(61k-23))}}}.....solve for{{{ k}}}

{{{(11k+17) =-(61k-23)}}}

{{{11k+17 =-61k+23}}}

{{{11k+61k =-17+23}}}

{{{72k =6}}}

{{{k =6/72}}}

{{{k =1/12}}}


your lines are:

line {{{l}}}

{{{y=(11k+17)x+1}}}

{{{y=(11(1/12)+17)x+1}}}

{{{y=(215/12)x+1}}}


line {{{l}}}’

{{{y=(1/(61k-23))x+2}}}

{{{y=(1/(61(1/12)-23))x+2}}}

{{{y=-(12/215)x+2}}}



{{{ graph( 600, 600, -5, 5, -5, 5, -(12/215)x+2, (215/12)x+1) }}}