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In the xy-coordinate plane, line L passes through the points (b, a) an

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In the xy-coordinate plane, line L passes through the points (b, a) an [#permalink]

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In the xy-coordinate plane, line L passes through the points (b, a) and (c, 0), and line M passes through the point (a, b) and the origin, where a, b, and c are different nonzero numbers. Do lines L and M intersect?

(1) \(a=(\frac{{\sqrt{5}-1}}{2})^b\)

(2) c < 0


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[Reveal] Spoiler: OA

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In the xy-coordinate plane, line L passes through the points (b, a) an [#permalink]

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Using definitions of lines on xy-plane we get
\(Y_1 = X*\frac{b}{a}\)
for Y2 we got this equation: \(\frac{(X-c)}{(b-c)} = \frac{Y}{a} => Y_2= \frac{a*X}{(b-c)}-\frac{c}{(b-c)}\).
It is fairly obvious that lines will intersect at the origin if C = 0, so lets try find out when lines are parallel.
Here we go, the condition for lines to be parallel is the equality of their constants near variable X, i.e
\(\frac{b}{a} = \frac{a}{(b-c)}\)
\(b^2 - b*c = a^2\)

using #1 we get
\(b^2 - b*c = b^2*(3 - \sqrt{5})/2\)
\(b - b*(3 - \sqrt{5})/2 = c\)
\(c = b*(1 - (3 - \sqrt{5})/2) = b*(\sqrt{5} - 1)/2 = a\)
which contradicts the conditions in our question where it says that a,b and c are different, thus lines are not parallel

Sufficient

#2 is clearly insufficient, just pick some random numbers \(a = 2, b = 1 (y_1 = \frac{x}{2})\), then \(y_2\) passes points \((c,0)\) and \((1,2)\). If \(c = -3, Y_2 = \frac{1}{2}*x + 1,5\), lines become parallel, and if \(c = -100\), they clearly intersect in the first quadrant

A

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Re: In the xy-coordinate plane, line L passes through the points (b, a) an [#permalink]

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New post 21 Feb 2016, 12:39
Bunel, can you please help explain the answer step by step.

I don't understand the following:
Using definitions of lines on xy-plane we get
Y 1 =X∗ba

for Y2 we got this equation: (X−c)/(b−c) =Y/a =>Y2 =a∗X/(b−c) −c/(b−c)


Thank you.

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Re: In the xy-coordinate plane, line L passes through the points (b, a) an [#permalink]

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New post 01 May 2016, 04:28
ashakil3 wrote:
Bunel, can you please help explain the answer step by step.

I don't understand the following:
Using definitions of lines on xy-plane we get
Y 1 =X∗ba

for Y2 we got this equation: (X−c)/(b−c) =Y/a =>Y2 =a∗X/(b−c) −c/(b−c)


Thank you.



two point line equation (a,b) and (c,d)

m = slope = d-b/c-a
equation will be y-b = (d-b/c-a)(x-a)
try to form the equation of one line from a,b and c,0 and other line from c,0 an 0.,0
you will get the above equations

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Re: In the xy-coordinate plane, line L passes through the points (b, a) an [#permalink]

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New post 02 May 2016, 07:13
ashakil3 wrote:
Bunel, can you please help explain the answer step by step.

I don't understand the following:
Using definitions of lines on xy-plane we get
Y 1 =X∗ba

for Y2 we got this equation: (X−c)/(b−c) =Y/a =>Y2 =a∗X/(b−c) −c/(b−c)


Thank you.


The standard equation of a line in x-y plane passing through points (a,b) and (c,d) :

\(\frac{y-b}{x-a} = \frac{d-b}{c-a}\).

Equation of line Y1 is created from the above equation of a line but now passing through the origin (0,0), substitute either (a,b) or (c,d) as (0,0) and you get (assuming (c,d) = (0,0))

\(\frac{y-b}{x-a} = \frac{d-b}{c-a}\) ---> \(\frac{y-b}{x-a} = \frac{0-b}{0-a}\) ---> \(\frac{y-b}{x-a} = \frac{b}{a}\)

---> After rearranging the terms, you get, \(y = x* (b/a)\)

For Y2, the OP has used the standard equation of a line (mentioned above).

Hope this helps.

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Re: In the xy-coordinate plane, line L passes through the points (b, a) an [#permalink]

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New post 29 Oct 2017, 00:12
Can anyone please explain how to derive b2−b∗c=b2∗(3−5√)/2?

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Re: In the xy-coordinate plane, line L passes through the points (b, a) an   [#permalink] 29 Oct 2017, 00:12
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