Question 936106
If the line passing through the points
({{{a}}}, {{{1}}}) and ({{{1}}}, {{{8}}})

 and parallel to the line passing through the points
({{{12}}}, {{{5}}}) and ({{{a + 2}}},{{{ 1}}})

to find equations, start with slope-intercept form and find a slope and y-intercept:

{{{y=mx+b}}}  use point ({{{1}}}, {{{8}}})

{{{8=m*1+b }}}

{{{8=m+b}}}

{{{m=8-b}}}............eq.1

for parallel line:

{{{y=mx+b}}}  use point ({{{12}}}, {{{5}}})

{{{5=m*12+b}}} 

{{{5=12m+b}}}

{{{12m=5-b}}}

{{{m=(5-b)/12}}}...........eq.2

the parallel line will have same slope: so, we have

{{{8-b=(5-b)/12}}}.....solve for {{{b}}}

{{{96-12b=5-b}}}

{{{96-5=12b-b}}}

{{{91=11b}}}

{{{b=91/11}}}

then slope of the first line is: 

{{{m=8-b}}} => {{{m=8-91/11=88/11-91/11=-3/11}}}

and equation of first line is:
 
{{{y=-(3/11)x+91/11}}} use point ({{{a}}}, {{{1}}}) to find {{{a}}}

{{{1=-(3/11)a+91/11}}} ..........solve for {{{a}}}

{{{(3/11)a=91/11-1}}}

{{{3a/11=91/11-11/11}}}

{{{3a/11=80/11}}} ...if denominators same then

{{{3a=80}}}

{{{a=80/3}}}
or
{{{a=26.66666666666667}}}

{{{a=26.7}}}=> the  point  is ({{{a}}}, {{{1}}})=({{{26.7}}}, {{{1}}})



find second equation:

slope is same {{{m =-3/11}}} find {{{b}}}
{{{-3/11 =(5-b)/12}}}
{{{-(3/11)12=5-b}}}
{{{b=36/11}}}


so, second equation is:

{{{y=-(3/11)x+3.27}}}  use point ({{{a + 2}}}, {{{1}}})

{{{1=-(3/11)(a + 2)+3.27}}}
{{{1-3.27=(-3/11)(a + 2)}}}
{{{-2.27/(-3/11)=a + 2}}}
{{{2.27/0.27=a + 2}}}

{{{8.4=a+2}}}
{{{a=8.4-2}}}
{{{a=6.4}}}  => the  point  is ({{{a + 2}}},{{{ 1}}})=({{{6.4 + 2}}}, {{{1}}})=({{{8.4}}}, {{{1}}})

{{{drawing( 600, 600, -10, 35, -10, 10,circle(26.7,1,.155),locate(26.7,1,p(26.7,1)),circle(8.4,1,.155),
locate(8.4,1,p(8.4,1)), graph( 600, 600, -10, 35, -10, 10, -(3/11)x+91/11, -(3/11)x+3.27)) }}}