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

Question

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.

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

(2) c < 0

Zhenek · Accepted Answer

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

ashakil3 · Answer

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.

sudhirmadaan · Answer

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

ENGRTOMBA2018 · Answer

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.

Rw27 · Answer

Can anyone please explain how to derive b2−b∗c=b2∗(3−5√)/2?

siddharth19 · Answer

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

GMATGuruNY · Answer

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.

A

mimishyu · Answer

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

(1)a= 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  ”
thus here we only have to examine “b/a ≠  ”(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 =    
transpose on both side --- b(b-c)=a^2
arrange the equation --- b^2=a^2+bc
substitute { b} for a
a^2 = { b}^2 = (b^2)*(1- c/b)
eliminate b^2 on both side
  ^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 = *b 
also as we already know a= *b 
a=c
….this totally contradict what the stimulus says “a, b, and c are different nonzero numbers”

thus if a= *b
b/a must not equal≠  
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

elPatron434 · Answer

I agree, I mistook b to be the exponent

bumpbot · Answer

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

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