Question 921315
an equation in slope-intercept form is: {{{y=mx+b}}} where {{{m}}} is a slope and {{{b}}} is y-intercept

given:

point ({{{1}}}, {{{-3}}})

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

to find: an equation of a line that passes through the given point and is parallel to the graph of the given equation


first write {{{y+2=4(x-1)}}} in slope-intercept form; we need to know what a slope is

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

{{{y=4x-4-2}}}

{{{y=highlight(4)x-6}}}...as you can see, the slope {{{m[1]=4}}}

you also know that parallel lines have {{{same}}}{{{ slope}}}; so, the line parallel to given line has slope {{{m[2]=4}}}

now, plug it in slope-intercept form: 

{{{y=highlight(4)x+b}}} , now use given point ({{{1}}}, {{{-3}}}) and substitute {{{1}}} for {{{x}}} and {{{-3}}} for {{{y}}}


{{{-3=4*1+b}}} .........solve for {{{b}}}

{{{-3=4*1+b}}}

{{{-3=4+b}}}

{{{-3-4=b}}}

{{{-7=b}}}

go back to {{{y=highlight(4)x+b}}}, plug in {{{b}}}

{{{highlight(y=4x-7)}}} ....this is an equation of a line parallel to {{{y+2=4(x-1)}}} and passes through the point ({{{1}}}, {{{-3}}})

here is the graph:

{{{drawing( 600, 600, -10, 10, -10, 10,circle(1,-3,.13),locate(1,-3,p(1,-3)), graph( 600, 600, -10, 10, -10, 10, 4x-6,4x-7)) }}}