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Parallel Lines

Expert replies
by Abdulla » Wed Nov 25, 2009 11:37 am
OA is C
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Abdulla
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Source: — Problem Solving |

by mp2437 » Wed Nov 25, 2009 12:46 pm
First, you should remember that parallel lines have the same slope! This is very important. Since they tell you that the product of the two lines is 9, you know that the slope of each line has to be 3 (3 *3 = 9). A line is of the form:

y = mx + b = 3x + b


We have two lines, y = 3x + b1 and y = 3x + b2. We are told of two points that are on two lines: (2,k) and (k,6) for each line. Plug it in:

y = 3x + b1 --> k = 3(2) +b1 --> k = 6 + b1
y = 3x + b2--> 6 = 3k + b2, or rearranging --> -3k = -6 + b2

Add the two equations: (k - 3k) = (6 +b1) + (-6 + b2) --> -2k = b1 + b2. We know that b1 + b2 = 4 (given), so -2k = 4, or k = -2. Plug this back into the two equations above to solve for b1 and b2:

(1) k = 6 + b1 --> -2 = 6 + b1, or b1 = -8
(2) 6 = 3k + b2 --> 6 = -6 + b2, or b2 = 12

so your lines are now:

(1) y = 3x - 8
(2) y = 3x + 12

Now plug the x values of each answer choice into these two equations to see if it matches up to the y-value. Choice C fits the bill with (1, -5), as you can see when you plug it into equation (1):

y = 3(1) - 8 = 3 - 8 = -5, so (1,-5) is a point on this line. Done!
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