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

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Math Expert
Joined: 02 Sep 2009
Posts: 52422
In the xy-coordinate plane, line L passes through the points (b, a) an  [#permalink]

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01 May 2015, 01:16
2
19
00:00

Difficulty:

95% (hard)

Question Stats:

30% (02:47) correct 70% (03:00) wrong based on 230 sessions

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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

Kudos for a correct solution.

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Joined: 17 Mar 2015
Posts: 116
In the xy-coordinate plane, line L passes through the points (b, a) an  [#permalink]

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03 May 2015, 08:10
2
4
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
##### General Discussion
Intern
Joined: 09 Jun 2015
Posts: 8
Re: In the xy-coordinate plane, line L passes through the points (b, a) an  [#permalink]

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21 Feb 2016, 11:39

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.
Manager
Joined: 29 Nov 2011
Posts: 96
Re: In the xy-coordinate plane, line L passes through the points (b, a) an  [#permalink]

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01 May 2016, 03:28
ashakil3 wrote:

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
CEO
Joined: 20 Mar 2014
Posts: 2636
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: In the xy-coordinate plane, line L passes through the points (b, a) an  [#permalink]

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02 May 2016, 06:13
ashakil3 wrote:

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.
Intern
Joined: 03 Sep 2015
Posts: 18
Re: In the xy-coordinate plane, line L passes through the points (b, a) an  [#permalink]

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28 Oct 2017, 23:12
Can anyone please explain how to derive b2−b∗c=b2∗(3−5√)/2?
Re: In the xy-coordinate plane, line L passes through the points (b, a) an &nbs [#permalink] 28 Oct 2017, 23:12
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