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Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
08 Nov 2012, 09:57

3

This post received KUDOS

Expert's post

smartass666 wrote:

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

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)

Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
04 Apr 2013, 09:15

3

This post received KUDOS

Expert's post

smartass666 wrote:

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

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.

Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
03 Apr 2013, 12:15

2

This post received KUDOS

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 _________________

Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
08 Nov 2012, 01:34

1

This post received KUDOS

smartass666 wrote:

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: 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.

Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
08 Nov 2012, 06:31

1

This post received KUDOS

(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

Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
02 Apr 2013, 09:25

1

This post received KUDOS

catennacio wrote:

(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

(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??

Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
02 Apr 2013, 15:39

1

This post received KUDOS

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)

Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
03 Apr 2013, 11:41

1

This post received KUDOS

Dipankar6435 wrote:

srcc25anu wrote:

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. )

Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
23 Mar 2014, 23:12

1

This post received KUDOS

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.

Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
02 Apr 2013, 20:01

srcc25anu wrote:

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

Re: If line L in the xy-coordinate plane has a positive slope, [#permalink]
03 Apr 2013, 20:29

hikaps14 wrote:

Dipankar6435 wrote:

srcc25anu wrote:

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.

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