Question 1129389
Find {{{k}}} so that the line through ({{{7}}},{{{2k}}}) and ({{{4}}},{{{-3}}}) is parallel to the line through ({{{1}}}, {{{k+1}}}) and ({{{3}}},{{{5}}}) 


use a slopes:

a slope of the line through ({{{7}}},{{{2k}}}) and ({{{4}}},{{{-3}}}) is:

{{{m=(-3-2k)/(4-7)}}}

{{{m=(-3-2k)/(-3)}}}

a slope of the parallel to the line that passes through (1, k+1) and (3,5) is:

{{{m[p]=(5-(k+1))/(3-1)}}}

{{{m[p]=(5-(k+1))/2}}}

{{{m[p]=(5-k-1))/2}}}

{{{m[p]=(4-k))/2}}}



since parallel lines have same slope, we have:

{{{m=m[p]}}}

{{{(-3-2k)/(-3)=(4-k)/2}}}....cross multiply

{{{2(-3-2k)=(-3)(4-k)}}}

{{{-6-4k=-12+3k}}}

{{{-6+12=4k+3k}}}

{{{6=7k}}}

{{{k=6/7}}}

so, slope is
{{{m=(-3-2(6/7))/(-3)}}}
{{{m=(-3-12/7)/(-3)}}}
{{{m=11/7}}}

same for parallel line
{{{m[p]=(11/7)}}}

then, points and slope for one line are:
 
 ({{{7}}},{{{2(6/7)}}}) and ({{{4}}},{{{-3}}}) 
=>({{{7}}},{{{12/7}}}) and ({{{4}}},{{{-3}}}) 



equation of this line is: use point-slope formula
{{{y-y[1]=m(x-x[1])}}}.......if  ({{{4}}},{{{-3}}}) 

{{{y-(-3)=(11/7)(x-4)}}}

{{{y+3=(11/7)x-(11/7)4}}}

{{{y=(11/7)x-44/7-3}}}

{{{y=(11/7)x-44/7-21/7}}}

{{{y=(11/7)x-65/7}}}-> one line


equation of the parallel line is: use point-slope formula

({{{1}}}, {{{6/7+1}}}) and ({{{3}}},{{{5}}}) 
=> ({{{1}}}, {{{13/7}}}) and ({{{3}}},{{{5}}}) 

{{{y-y[1]=m(x-x[1])}}}.......if {{{m[p]=(11/7)}}}and ({{{3}}},{{{5}}}) 

{{{y-5=(11/7)(x-3)}}}

{{{y-5=(11/7)x-(11/7)3}}}

{{{y-5=(11/7)x-33/7}}}

{{{y=(11/7)x-33/7+5}}}

{{{y=(11/7)x-33/7+35/7}}}

{{{y=(11/7)x+2/7}}}-> parallel line


=>({{{7}}},{{{12/7}}}) and ({{{4}}},{{{-3}}}) 
=> ({{{1}}}, {{{13/7}}}) and ({{{3}}},{{{5}}}) 

{{{drawing( 600, 600, -15, 15, -15, 15,
circle(7,12/7,.12),circle(4,-3,.12),
locate(7,12/7,p(7,12/7)),locate(4,-3,p(4,-3)),
circle(1,13/7,.12),circle(3,5,.12),
locate(1,13/7,p(1,13/7)),locate(3,5,p(3,5)),
 graph( 600, 600, -15, 15, -15, 15, (11/7)x-65/7, (11/7)x+2/7)) }}}