Question 211448
You must learn some facts about lines, their graphs
and their equations.
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(1). The equation of a non-vertical graphed line that goes through point
(x<sub>1</sub>,y<sub>1</sub>) and has slope {{{m}}} is

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

(3). The equation of a non-vertical graphed line that has slope {{{m}}} and 
y-intercept {{{"(0,b)"}}} is 

{{{y=mx+b}}}

(4). Two non-vertical parallel lines have the same slope.

(5). Two non-vertical perpendicular lines have slopes one of which
is the reciprocal of the other with the sign changed.

(6). A vertical line has no slope and has the equation {{{x = a}}} where
{{{a}}} represents its distance from the y-axis, right of the y-axis
if {{{a}}} is positive and left of the y-axis if {{{a}}} is negative.
{{{x=0}}} is the equation of the vertical line that coincides with
the y-axis.

(7). A horizontal line has slope {{{m=0}}}.

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write an equation for each line, then graph the line.
<pre><font size = 4 color = "indigo"><b>
through (-1,3) and parallel to {{{y=2x+1}}}

We draw the line {{{y=2x+1}}} by plotting a couple of points
and draw it in light green:

{{{drawing(200,200,-5,5,-5,5,

graph(200,200,-5,5,-5,5,7,2x+1)  )}}}

Next we plot the point (-1,3)

{{{drawing(200,200,-5,5,-5,5,locate(-3.4,3.41,"(-1,3)"),

line(-1+.1,3,-1-.1,3), line(-1,3+.1,-1,3-.1), line(-1+.1,3+.1,-1-.1,3-.1), line(-1-.1,3-.1,-1+.1,3+.1), 

graph(200,200,-5,5,-5,5,7,2x+1)  )}}}

And we draw a black line through that point 
parallel to the green line:

{{{drawing(200,200,-5,5,-5,5, line(-5,-5,4,13),
locate(-3.4,3.41,"(-1,3)"),
line(-1+.1,3,-1-.1,3), line(-1,3+.1,-1,3-.1), line(-1+.1,3+.1,-1-.1,3-.1), line(-1-.1,3-.1,-1+.1,3+.1), 

graph(200,200,-5,5,-5,5,7,2x+1)   )}}}

That black line is the line we need the equation for.

So, from {{{y=2x+1}}} we use (3) above to find that {{{m=2}}}

From (4) we know that we will use the same slope {{{m=2}}}.

Next we use (1) to find the equation:

{{{y-y[1]=m(x-x[1])}}} where (x<sub>1</sub>,y<sub>1</sub>) = (-1,3)
and {{{m=2}}}

{{{y-3=2(x-(-1))}}}
 
{{{y-3=2(x+1)}}}

{{{y-3=2x+2}}}

{{{y=2x+5}}}

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through (2,2) and perpendicular to {{{y=(-3/5)x+2}}}
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We draw the line {{{y=(-2/3)x+2}}} by plotting a couple of points
and draw it in light green:

{{{drawing(200,200,-5,5,-5,5,

graph(200,200,-5,5,-5,5,7,(-3/5)x+2)  )}}}

Next we plot the point (2,2)

{{{drawing(200,200,-5,5,-5,5,locate(2,2,"(2,2)"),

line(2+.1,2,2-.1,2), line(2,2+.1,2,2-.1), line(2+.1,2+.1,2-.1,2-.1), line(2-.1,2-.1,2+.1,2+.1), 

graph(200,200,-5,5,-5,5,7,(-3/5)x+2)  )}}}

And we draw a black line through that point 
perpendicular to the green line:

{{{drawing(200,200,-5,5,-5,5, line(8,12,-7,-13),
locate(2,2,"(2,2)"),
line(2+.1,2,2-.1,2), line(2,2+.1,2,2-.1), line(2+.1,2+.1,2-.1,2-.1), line(2-.1,2-.1,2+.1,2+.1), 

graph(200,200,-5,5,-5,5,7,(-3/5)x+2)   )}}}

That black line is the line we need the equation for.

So, from {{{y=(-3/5)x+2}}} we use (3) above to find that {{{m=-3/5}}}

From (5) we know that we will use the slope which is found
by inverting {{{-3/5}}} and changing its sign, and so we
use {{{m=5/3}}} for the black line.

Next we use (1) to find the equation:

{{{y-y[1]=m(x-x[1])}}} where (x<sub>1</sub>,y<sub>1</sub>) = (2,2)
and {{{m=5/3}}}

{{{y-2=(5/3)(x-2)}}}
 
{{{y-2=(5/3)(x-2)}}}

{{{y-2=(5/3)x-10/3}}}

{{{y=(5/3)x-10/3+2}}}

{{{y=(5/3)x-10/3+6/3}}}

{{{y=(5/3)x-4/3}}}


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through (-3,4) and vertical
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We plot the point:

{{{drawing(200,200,-5,5,-5,5,locate(-2.9,4.5,"(-3,4)"),

line(-3+.1,4,-3-.1,4), line(-3,4+.1,-3,4-.1), line(-3+.1,4+.1,-3-.1,4-.1), line(-3-.1,4-.1,-3+.1,4+.1), 

graph(200,200,-5,5,-5,5)  )}}}

and draw a vertical line through it:

{{{drawing(200,200,-5,5,-5,5,locate(-2.9,4.5,"(-3,4)"),
line(-3,-6,-3,6),
line(-3+.1,4,-3-.1,4), line(-3,4+.1,-3,4-.1), line(-3+.1,4+.1,-3-.1,4-.1), line(-3-.1,4-.1,-3+.1,4+.1), 

graph(200,200,-5,5,-5,5)  )}}}

Now we use (6) above and realize that the vertical line is 3
units to the left of the the y-axis and thuse its equation
is {{{x=-3}}}

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through (4,1) and horizontal
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We plot the point:

{{{drawing(200,200,-5,5,-5,5,locate(2.9,2.1,"(4,1)"),

line(4+.1,1,4-.1,1), line(4,1+.1,4,1-.1), line(4+.1,1+.1,4-.1,1-.1), line(4-.1,1-.1,4+.1,1+.1), 

graph(200,200,-5,5,-5,5)  )}}}

and draw a horizontal line through it:

{{{drawing(200,200,-5,5,-5,5,locate(2.9,2.1,"(4,1)"),

line(4+.1,1,4-.1,1), line(4,1+.1,4,1-.1), line(4+.1,1+.1,4-.1,1-.1), line(4-.1,1-.1,4+.1,1+.1), line(-6,1,6,1),

graph(200,200,-5,5,-5,5)  )}}}

Now we use (7) above which tells us the slope {{{m=0}}}

Next we use (1) to find the equation:

{{{y-y[1]=m(x-x[1])}}} where (x<sub>1</sub>,y<sub>1</sub>) = (4,1)
and {{{m=0}}}

{{{y-1=0(x-4)}}}
 
{{{y-1=0}}}

{{{y=1}}}


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Edwin</pre>