If line L in the xy-coordinate plane has a positive slope,

Dipankar,

To understand why stmt 1 is sufficient, let's revisit a line equation. A line is represented using y = mx +k in which m is the slope and k is the y-intercept.

Taking stmt 1, the m = d-b/c-a = d/c. Hence y = (d/c )x + k and to find k just use one of the points on the line. Using point (c, d) in the line y= (d/c)x + k gives k = 0. hence the equation of the line passing through points (a,b) and (c,d) is y = (d/c)x and to find x intercept set y = 0, which will result in x = 0.

Hope this is clear..

//kudos please, if this explanation helps you

Responding to a pm:

The equation of a line is
y = mx + c
We are given that m is positive. We need to find the x intercept.


(1) There are different points (a, b) and (c, d) on line L such that ad = bc.

So d/c = b/a = y co-ordinate/x co-ordinate

Equation of the line: y/x = m + c/x

So the value of y/x is dependent on the value of x and hence will change for every point on the line.
If the value is the same for two distinct points, it means c = 0. So the equation of the line is
y = mx and it passes through (0, 0) and has x intercept as 0.
Sufficient

(2) There are constants m and n such that the points (m, n) and (–m, –n) are both on line L
Again, since -n/-m = n/m, we have that y/x is same for two distinct points.
So by the same logic as above, this statement is also sufficient alone.

Answer (D)

Statement 1: ad =bc =>
b/a = d/c = m (assume)
Each point lies on a line which passes through origin. (y=mx+c and c=0 making it y=mx)

So x intercept =0 . Sufficient

Statement 2: (m,n) and (-m,-n) on same line. m and -m are mirror image across Y axis. n and -n are mirror image across x axis. thus (m,n) and (-m,-n) are mirror image of each other across origin. Thus a line joining these points must go through origin.

So x intercept =0 . Sufficient

Hence ans D it is.

(1): a/b = c/d --> 2 points pass through the same slope. Note that k slope = y/x (y advances a units for for every unit x advances) --> line passes through 0 (y=kx) --> suff
(2) (m,n) and (-m, -n) belong to (l) --> they are opposite of each other through the origin O --> suff

--> D

Given fact: Line L has a positive slope.

Statement 1) ad=bc, where(a,b) is different from (b,c)
Now if you see the diagram, you will observe that the product of two "outer" values is greater than the product of two "inner"values.
The first diagram clearly depicts a line passing through quadrants III, II and I , and is also a representative of other such lines. If you clearly see, the product of and d will only be equal to the product of b and c, only when they are either sides of origin and opposite to each other. (as shown in the other diagram)

Statement 2) states the same in other words.

Hence D

(1): a/b = c/d --> 2 points pass through the same slope. Note that k slope = y/x (y advances a units for for every unit x advances) --> line passes through 0 (y=kx) --> suff

Can some mathwiz clarify a little more on this.
I do not understand how if a/b = c/d the line passes through the origin??

ST 1: ad = cb for any 2 combination of points (a,b) and (c,d) that fulfils the equation ad = bc will pass through origin (as shown in attached graph)
hence SUFFICIENT (as x intercept = 0)

St 2: (m,n) and (-m, -n) lie on the line. Any combination of points on a line that satisfy this equation will pass through origin
hence SUFFICIENT (as x intercept = 0)

Ans is D

With the risk of sounding especially dumb i wish to resubmit my doubts regarding this
St 2 does make sense to me (m,n) and (-m,-n) are mirror images of each other about the origin. Thus they must lie on line y = x . GOT IT
St 1 still unclear
srcc25anu your graph is pretty neat but i just cannot comprehend how knowing that the product of x and y coordinates being equal (i.e x1*y1 = x2*y2) makes the line containing them pass though the origin
Can i get a link to some articles where such nuances of coordinate geometry are explained in detail
Apologies for being such a pain

Just to keep you correct on ST2. The Line is Y=KX not Y=X. The slope may or may not be 1.

In case of ST2.

Slope of any line = (Y2 - Y1)/X2- X1
if point (X1,Y1) is origin then slope = Y2/X2

Similary, for any other point (X3,Y3) on same line

Slope = Y3/X3

Y3/X3 = Y2/X2
X2 * Y3 = Y2 * X3 ( This holds true for all lines passing through origin)

for points (a,b) & (c,d)

a * d = b * c ( This is what was given in question. )

Does this help ?

It helps greatly. Thanks
So I can add this to my cheat sheet now-
If y1/x1 = y2/x2 then the line containing the points (x1,y1) and (x2,y2) passes through the origin.

From F.S 1, we have that the slope of the given line is (d-b)/(c-a). Again, we have d = (bc)/a ; the slope is = b/a. Now in the equation, y = mx+c, just plugin the value of m = b/a and the point (a,b)--> we get c= 0, thus the line passes through the origin. The x intercept is 0.Sufficient.

From F.S 2, we have that both the points (m,n) and (-m,-n) lie on the same line.Thus, we can see that the mid-point of these 2 points is at the origin. Line passes through the origin, hence the x-intercept = 0. Sufficient.

D.
_________________

Hi Honchos,
Let me try to explain:
as you know eqn of line is y = mx + e (e is y intercept)

St1: Points (a,b) and (c,d) lies on above line.
So
b = ma + e ..(i)
d = mc + e ..(ii)
dividing (i) by (ii) we get
b/d = (ma + e) / (mc + e).
Also we know ad = bc. Thus b/d = a/c. Substituting in above we get:
a/c = (ma + e) / (mc + e)
(mac + ea) = (mac + ec)
solving above, ea - ec = 0 --> e(a-c) = 0, from this we can say that either e = 0 or a = c. But a cannot be equal to c (as they are two different points). So e = 0.
Final eqn; y = mx .. which indicates that line passes through origin. So x intercept = 0.

Sufficient.

st2: since (m,n) & (-m,-n) are two end points of segment passing through origin, and line l also contains this points, we can say that line l passes through origin. So x intercept = 0.
Sufficient.

Answer: D

Hope this helps!

This looks like a humugously crazy problem but a little analysis of the concepts will help you reach the solution in 1:15 sec (The time I took )
First lets remove all the flab and look at the core of the question .
y=mx+c
x intercept so y=0 hence x=-c/m

So we need the value of C and the slope of the line.

Statement1- it is given (a,b) and (c,d) and ad=bc
so manipulating the factors a/b=c/d=k . the ratio of all values of x coordinates and y coordinates will be the same only when the line passes throught the origin. x=y (equation of the line) . hence when y=0 then x=0
then Statemnt 1 is sufficient.

This can be proved algebraically too, let a=cb/d. now put the values in the equation of line y-y1=m(x-x1)

Statement 2: We have two point (m,n) and (-m,-n) and they both lie of the line L. Wow what will be the midpoint (0,0). Voila! you got it. The line passes through the origin. What else can you ask for. We have x=0 when y=0.
Sufficienet

Thus D.

Treat yourself with a cupcake now

There is a pretty convenient shortcut we can apply to these problems

any time between two points x1/y1 = x2/y2 then line must pass through the origin

Statement 1

b/a = c/d - the rule applies

Suff

Statement 2

m/n = -m/-n *just remember two negatives in a fraction are positive

Suff

Thus

"D"

Statement 1:
Let (a,b) = (2,3) and (c,d) = (4,6).
Now midpoint of these 2 points = (3,4.5)
Hmm... so with each 'step down' of x by 1, y decreases by 1.5.
Line moves from (4,6) ... to (3,4.5)... to (2,3)... to (1,1.5) ... to (0,0) !
So... for x = 0, y = 0.
Hence... zero x-intercept.
Sufficient.

Statement 2:
Let m = 1 and n = 1.
Hence... points are (1,1) and (-1,-1).
Definitely, line passes through origin.
Sufficient.

Answer: D

OFFICIAL EXPLANATION:

If line L in the xy-coordinate plane has a positive slope, what is the x-intercept of L ?

(1) There are different points (a, b) and (c, d) on line L such that ad = bc.

(2) There are constants m and n such that the points (m, n) and (–m, –n) are both on line L.

STATEMENT (1) SUFFICIENT: The easiest way to deal with this statement is to pick smart numbers, and see what the x-intercept is in each case. Plug in sets of numbers a, b, c, and d such that ad = bc. In every case, the x-intercept of the resulting line will be 0; plug in enough points to become convinced of this pattern.

For instance, let a = 6, b = 4, c = 3, d = 2. Then the line passes through the points (6, 4) and (3, 2). By finding the equation of the line we can determine that the x-intercept of the line is (0, 0).

Try another set of numbers: a = –10, b = –5, c = –2, d = –1. In this case, the line passes through (–10, –5) and (–2, –1). Again, by finding the equation of the line, we can determine that the x-intercept of the line is (0, 0).

Continue plugging until you are convinced that the pattern won’t have any exceptions. The x-intercept is always 0; the statement is sufficient.

Algebra:

Solve for one of the variables in the equation ad = bc, say d:
\(d = \frac{bc}{a}\)

Therefore, line L passes through the points \(( a, b) and (c, \frac{bc}{a} )\)
To write an equation of the line, find its slope: (y2-y1)/(x2-x1)=
\([(\frac{bc}{a} -b)] / [(c - a)]\)

Multiply the top and bottom of this fraction by a to eliminate the smaller denominator, giving \( \frac{[(bc - ba)]}{[a(c-a)]} = \frac{[b(c-a)]}{[a(c-a)]} = \frac{b}{a}\)

The slope of the line is therefore \frac{b}{ a}. So, the equation of the line can be written in point-slope form, using the point ( a, b) and the slope, as
\( y-b = \frac{b}{a }(x-a)\).
Expanding this equation gives, \(y-b = (\frac{b}{a})x -b\) or just \(y = \frac{b}{a}x\).
Plug y = 0 to find the x-intercept, giving an x-intercept of 0. The statement is sufficient.




STATEMENT (2) SUFFICIENT: There are two cases to consider:

First, we might have m = n = 0. In this case, the point (0, 0) is on line L, so that point is the x-intercept.

Second, we might have any other numbers m and n. In this case, ( m, n) and (– m, – n) are two different points, so their midpoint (which is distinct from either of them) must also be on the line.

The midpoint of ( m, n) and (– m, – n) is (0, 0), so the x-intercept of the line must be 0.

You can also solve the second case by solving for the equation of the line.
If the line passes through two different points ( m, n) and (– m, – n), then standard algebra techniques yield the equation \(y = \frac{m}{n} x\) for the line.
Plug y = 0 to find the x-intercept, revealing an x-intercept of (0, 0).

In each case the x-intercept is 0, so the statement is sufficient.

@Brunnel if (m,n) and (-m,-n) are on the line then the line must pass through origin, so why is B not satisfactory?

The key to solving the equation is getting all the variable sorted out and having the value of the slope and constant extracted keeping that mind

(1) There are different points (a, b) and (c, d) on line L such that ad = bc.

Since we have 2 equations and 2 variables since y- can be eleminated from the frame we get the value x

Hence suff
(2) There are constants m and n such that the points (m, n) and (–m, –n) are both on line L

Exactly the same reasoning all that we need to get is x=-c/m

m=-2
and with one substitution we get the value of C
Hence suff
Therefore IMO D

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________