Question 81763
For lines to be perpendicular their slopes have to be negative inversions of each other.
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Examples: 
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If one line has a slope of +8 the other line must have a slope of -1/8 for them to be 
perpendicular.
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If one line has a slope of -6 the other line must have a slope of +1/6 for them to be 
perpendicular.
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If one line has a slope of +1/4 the other line must have a slope of -4 for them to be 
perpendicular.
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If one line has a slope of -1/2 the other line must have a slope of +2 for them to be 
perpendicular.
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As you can see, the rule is (in non-math terms) given one slope, flip it over and change
the sign" and that will give you the slope of the other.
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Given a slope of 10 (or 10/1) flip it over to get 1/10 and change the sign to minus
to get the slope of the perpendicular as -1/10.
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Now to your problem ... all the equations you were given are in the slope intercept 
form ... y = mx + b.  m, the multiplier of x is the slope. So let's look at the slope for
each equation:
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a) y=(5/3)x-3
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The slope in this equation is (5/3). To find the slope of the line that is perpendicular
to this line, flip the slope to 3/5 and change the sign to minus.  A perpendicular
line will therefore, have a slope of -3/5.  None of the other equations has that slope.

b) y=3x-5/3
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In this equation, the slope of the graph (that is, the multiplier of x is +3. To find the
slope of a line perpendicular to it, flip it over to 1/3 and change the sign to minus.
The slope of a perpendicular line will therefore be -1/3.  Notice that equation c) has
this slope. This tells you that the graph of equation c) is a line that is perpendicular
to the graph of equation b). But let's work the remaining two equations, just for practice.
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c) y=-1/3x+5/3
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The slope of the graph of this equation is (according to its equation) -1/3.  Therefore,
the graph of a line perpendicular to it will have a slope of (flip and change signs)
+3/1 or just +3.  The graph of equation b) has that slope ... which we already knew.
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d) y=1/3x-5/3
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The graph of this equation has a slope of 1/3, so the graph of a line that is perpendicular
must have (flip and change signs) a slope of -3/1 or just -3.  None of the other three
equations has this slope. b) comes close, but the sign of the slope in b) is plus, not
minus 3 so it is not close to being perpendicular.
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Hope this helps you to understand the problem and how to work it.  The answer is again,
that the graphs of equations b) and c) are the perpendicular pair.
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