Question 374881
1. Find the equation of the line that passes through the point of intersection of -2x + 4y = 14 and 5x - 3y = -14 and is perpendicular to 2x - 3y + 6 = 0. 
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First of all find the point of intersection by solving the system, x & y
-2x + 4y = 14
+5x - 3y = -14
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Multiply the 1st equation by 5, the 2nd equation by 2, results:
-10x + 20y = 70
+10x - 6y = -28
------------------adding eliminates x, find y
14y = 42
y = {{{42/14}}}
y = 3
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Find x using the 2nd equation: +5x - 3y = -14; replace y with 3
5x - 3(3) = -14
5x - 9 = -14
5x = -14 + 9
5x = -5
x = -1
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The point of intersection: x=-1, y=3
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We want the equation of a line that is perpendicular to 2x - 3y + 6 = 0
and pass thru the point x=-1, y=3
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Find the slope of the equation 2x - 3y + 6 = 0, put it in the slope/intercept form, namely: y = mx + b
2x - 3y + 6 = 0
subtract 2x and 6 from both sides
-3y = -2x - 6
Y has to be positive, mult by -1
3y = 2x + 6
Divide both sides by 3
y = {{{2/3}}}x + 2
The slope: m1 = {{{2/3}}} 
:
The relationship of the slopes of perpendicular lines is m1*m2 = -1
we know m1, find m2
{{{2/3}}}*m2 = -1
divide both sides by 3/2
m2 = -1*{{{3/2}}}
m2 = {{{-3/2}}}
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Using the point/slope form: y - y1 = m(x - x1)
x1=-1; y1=3 (the point of intersection)
m = {{{-3/2}}}
so we have:
y - 3 = {{{-3/2}}}[x - (-1)]
y - 3 = {{{-3/2}}}[x + 1]
y - 3 = {{{-3/2}}}x - {{{3/2}}}
add 3 to both sides
y = {{{-3/2}}}x - {{{3/2}}} + 3
which is
y = {{{-3/2}}}x - {{{3/2}}} + {{{6/2}}}
y = {{{-3/2}}}x + {{{3/2}}}
Put in the standard form, multiply by 2
2y = -3x + 3
Add 3x to both sides
3x + 2y = 3; is the equation of a line that passes  thru the intersecting point -1,3
and is perpendicular to the line 2x - 3y + 6 = 0.