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In the xy-coordinate plane, line L passes through the points (b, a) an 01 May 2015, 02:16
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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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In the xy-coordinate plane, line L passes through the points (b, a) an 03 May 2015, 09:10
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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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In the xy-coordinate plane, line L passes through the points (b, a) an 21 Feb 2016, 12:39
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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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In the xy-coordinate plane, line L passes through the points (b, a) an 01 May 2016, 04:28
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ashakil3
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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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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In the xy-coordinate plane, line L passes through the points (b, a) an 02 May 2016, 07:13
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ashakil3
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.
Show more

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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In the xy-coordinate plane, line L passes through the points (b, a) an 29 Oct 2017, 00:12
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Can anyone please explain how to derive b2−b∗c=b2∗(3−5√)/2?
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In the xy-coordinate plane, line L passes through the points (b, a) an 09 Mar 2019, 06:16
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Zhenek
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

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hi zehnek , Bunuel , i understand all the steps of above solution , but if stat (1) contradicts the condition given in question of all a,b,c being different, how does that imply lines are not parallel ?
thanks in advance
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In the xy-coordinate plane, line L passes through the points (b, a) an 24 Apr 2020, 10:45
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Bunuel
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
Show more

Lines L and M will intersect if they are not parallel.
Parallel lines have EQUAL SLOPES.
Thus, the answer to the question stem will be NO if Lines L and M have equal slopes.
Question stem, rephrased:
Do Lines L and M have equal slopes?

Statement 1:
Let \(b=2\), with the result that \(a=\sqrt{5}-1\)
Since Line L passes through \((2, \sqrt{5}-1)\) and \((c, 0)\), the slope of Line L = \(\frac{\sqrt{5}-1-0}{2-c} = \frac{\sqrt{5}-1}{2-c}\)

Since Line M passes through \((\sqrt{5}-1, 2)\) and \((0, 0)\), the slope of Line M = \(\frac{2-0}{\sqrt{5}-1-0} = \frac{2}{\sqrt{5}-1}\)

If the two slopes are equal, we get:
\(\frac{\sqrt{5}-1}{2-c}=\frac{2}{\sqrt{5}-1}\)
\((\sqrt{5}-1)^2 = 2(2-c)\)
\(5+1-2\sqrt{5}=4-2c\)
\(2c=2\sqrt{5}-2\)
\(c=\sqrt{5}-1=a\)
Not viable, since the prompt requires that a and c be DIFFERENT nonzero numbers.

Implication:
Since it is not viable for L and M to have equal slopes, they must have DIFFERENT slopes.
Thus, the answer to the rephrased question stem is NO.
SUFFICIENT.

Statement 2:
No information about a or b.
INSUFFICIENT.

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In the xy-coordinate plane, line L passes through the points (b, a) an 24 Apr 2020, 22:56
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Bunuel
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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for (1) a=[(√5-1)/2]^b this is a mistake, it should write a=[(√5-1)/2]*b, or else the calculation to this question
will be more complicated since it will have to do with logarithm or exponential

(1)a=[(√5-1)/2]b


as we know that if two lines intersect, they must not parallel each other
and if lineL & lineM parallel, according to formula of slope “b/a must equal [a/(b-c)]”
thus here we only have to examine “b/a ≠ [a/(b-c)]”(true or not) in order for us to prove lines L and M intersect

if we want to examine whether under (1) the above equation true or not
first we assume b/a = [a/(b-c)]
transpose on both side --- b(b-c)=a^2
arrange the equation --- b^2=a^2+bc
substitute {[(√5-1)/2]b} for a
a^2 = {[(√5-1)/2]b}^2 = (b^2)*(1- c/b)
eliminate b^2 on both side
[(√5-1)/2] ^2 = 1- c/b
(5-2√5+1)/4 = 1- c/b
c/b = 1 - (6-2√5)/4 = 1 - (3-√5)/2 =(√5-1)/2
c =[(√5-1)/2]*b
also as we already know a=[(√5-1)/2]*b
a=c
….this totally contradict what the stimulus says “a, b, and c are different nonzero numbers”

thus if a=[(√5-1)/2]*b
b/a must not equal≠ [a/(b-c)]
lines L and M not parallel, so they must intersect
sufficient



(2) c < 0

I draw coordinate in attachment to elaborate as for why (2) is insufficient
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In the xy-coordinate plane, line L passes through the points (b, a) an 06 May 2020, 05:56
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mimishyu
Bunuel
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.


for (1) a=[(√5-1)/2]^b this is a mistake, it should write a=[(√5-1)/2]*b, or else the calculation to this question
will be more complicated since it will have to do with logarithm or exponential

(1)a=[(√5-1)/2]b


as we know that if two lines intersect, they must not parallel each other
and if lineL & lineM parallel, according to formula of slope “b/a must equal [a/(b-c)]”
thus here we only have to examine “b/a ≠ [a/(b-c)]”(true or not) in order for us to prove lines L and M intersect

if we want to examine whether under (1) the above equation true or not
first we assume b/a = [a/(b-c)]
transpose on both side --- b(b-c)=a^2
arrange the equation --- b^2=a^2+bc
substitute {[(√5-1)/2]b} for a
a^2 = {[(√5-1)/2]b}^2 = (b^2)*(1- c/b)
eliminate b^2 on both side
[(√5-1)/2] ^2 = 1- c/b
(5-2√5+1)/4 = 1- c/b
c/b = 1 - (6-2√5)/4 = 1 - (3-√5)/2 =(√5-1)/2
c =[(√5-1)/2]*b
also as we already know a=[(√5-1)/2]*b
a=c
….this totally contradict what the stimulus says “a, b, and c are different nonzero numbers”

thus if a=[(√5-1)/2]*b
b/a must not equal≠ [a/(b-c)]
lines L and M not parallel, so they must intersect
sufficient



(2) c < 0

I draw coordinate in attachment to elaborate as for why (2) is insufficient
Show more

I agree, I mistook b to be the exponent
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In the xy-coordinate plane, line L passes through the points (b, a) an 26 Feb 2024, 09:05
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