Question 1115672


1.

equation of the line is

{{{y=mx+b}}} where {{{m}}} is a slope and {{{b}}} y-intercept
and slope point equation is:
{{{y-y[1]=m(x-x[1])}}}

choose two points from the given graph:

({{{1}}},{{{1}}}) and  ({{{-1}}},{{{0}}}) 

find slope:


{{{m=(y[2]-y[1])/(x[2]-x[1])}}} 


{{{m=(0-1)/(-1-1)}}} 

{{{m=-1/(-2)}}} 

{{{m=1/2}}} 

use slope point formula:

{{{y-y[1]=m(x-x[1])}}}.....plug in {{{m=1/2}}}  and coordinates of ({{{-1}}},{{{0}}}) 

{{{y-0=(1/2)(x-(-1))}}}

{{{y=(1/2)(x+1)}}}

{{{y=(1/2)x+1/2}}} -> your equation



2.
given: gradient {{{m=1/2}}}  and ({{{4}}},{{{-2}}}) 


use slope point formula:

{{{y-y[1]=m(x-x[1])}}}

{{{y-(-2)=(1/2)(x-4)}}}

{{{y+2=(1/2)x-4/2}}}

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

{{{y=(1/2)x-2-2}}}

{{{y=(1/2)x-4}}}-> your equation


3.

given: 
line passes through the point  ({{{-1}}},{{{2}}}) 
and is parallel to the line {{{y= x+4}}}

recall definition: parallel lines have {{{same}}} slope

since the line {{{y= x+4}}} has a slope {{{m=1}}},  parallel  line has a slope {{{m=1}}}

use slope point formula:

{{{y-y[1]=m(x-x[1])}}} and  the point  ({{{-1}}},{{{2}}}) 

{{{y-2=(1)(x-(-1))}}}

{{{y-2=x+1}}}

{{{y=x+1+2}}}

{{{y=x+3}}}->the line parallel to the line {{{y= x+4}}} and passes through the point  ({{{-1}}},{{{2}}}) 


4. this link is same as in # 1.