Question 714460
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Find an equation of the line that passes through the point (-1,3) and is parallel to the line passing through the points (-2,-3) and (2,5).

{{{drawing(400,400,-7,7,-7,7,graph(400,400,-7,7,-7,7), circle(-2,-3,.1),circle(2,5,.1),  locate(-3,3.2,"(-1,3)"), 
locate(-4.2,-2.8,"(-2,-3)"), locate(2,5,"(2,5)"),circle(-1,3,.1), red(line(-11,-21,13,27)),green(line(-12,-19,14,33)) )}}}
<pre>
We want to find the equation of the green line through (-1,3).  
We know that it has the SAME slope as the red line thru (-2,-3) and (2,5).

So we find the slope of the red line using the slope formula:

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

m = {{{(5-(-3))/(2-(-2))}}} = {{{(5+3)/(2+2)}}} = {{{8/4}}} = 2

So the slope of the green line is also 2.

Next we use the point-slope formula:

y - y<sub>1</sub> = m(x - x<sub>1</sub>) where (x<sub>1</sub>,y<sub>1</sub>) = (-1,3)

y - 3 = 2(x - (-1))
y - 3 = 2(x + 1)
y - 3 = 2x + 2
    y = 2x + 5

That's it.

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</pre>
Find an equation of the line that passes through the point (1,-2) and is perpendicular to the line passing through the point (-2,-1) and (4,3).

{{{drawing(400,400,-7,7,-7,7,graph(400,400,-7,7,-7,7), circle(-2,-1,.1),circle(4,3,.1),  locate(1,-2,"(1,-2)"), 
locate(-2,-1,"(-2,-1)"), locate(4,3,"(4,3)"),circle(1,-2,.1), red(line(-17,-11,13,9)),green(line(-19,28,13,-20)) )}}}
<pre>
We want to find the equation of the green line through (1,-2).  
We know that it has the NEGATIVE RECIPROCAL of the slope of the red line 
thru (-2,-1) and (4,3).

So we find the slope of the red line using the slope formula:

m = {{{(y[2]-y[1])/(x[2]-x[1])}}}
where (x<sub>1</sub>,y<sub>1</sub>) = (-2,-1)
and where (x<sub>2</sub>,y<sub>2</sub>) = (4,3)

m = {{{(3-(-1))/(4-(-2))}}} = {{{(3+1)/(4+2)}}} = {{{4/6}}} = {{{2/3}}}

So the slope of the green line is the NEGATIVE RECIPROCAL of {{{2/3}}},
which is {{{-3/2}}} .

Next we use the point-slope formula:

y - y<sub>1</sub> = m(x - x<sub>1</sub>) where (x<sub>1</sub>,y<sub>1</sub>) = (1,-2)

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

Clear the fraction by multiplying through by 2

2y + 4 = -3(x - 1)
2y + 4 = -3x + 3
2y = -3x - 1

Divide through by 2

 y = {{{-3/2}}}x - {{{1/2}}}

That's it.

PARALLEL LINES HAVE THE SAME SLOPE.

PERPENDICULAR LINES HAVE SLOPES WHICH ARE THE
NEGATIVE RECIPROCALS OF EACH OTHER.

Edwin</pre>