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lie at a unit

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lie at a unit

by sanju09 » Wed Mar 10, 2010 5:33 am
How many points on the line x + y = 4 are there that lie at a unit distance from the line 4 x + 3 y = 10?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
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by kstv » Wed Mar 10, 2010 6:33 am
The two st. lines are not parallel. They will meet at a point. So there are 4 points of either side of the intersection when the distance will be 1 unit between them.
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by firdaus117 » Wed Mar 10, 2010 7:47 am
Option Bkstv has already explained it.
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by hooliganpete » Wed Mar 10, 2010 4:19 pm
The two st. lines are not parallel. They will meet at a point. So there are 4 points of either side of the intersection when the distance will be 1 unit between them.
Option B kstv has already explained it.

I may have missed something but kstv says there are 4 points and firdasu117 says there are 2. Would someone mind breaking this down just a little more?

Thanks!
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by kstv » Wed Mar 10, 2010 11:56 pm
x + y = 4 and 4 x + 3 y = 10 are not parallel nor are they intersecting at 90°. So there should be 4 points.
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by firdaus117 » Thu Mar 11, 2010 12:10 am
kstv wrote:x + y = 4 and 4 x + 3 y = 10 are not parallel nor are they intersecting at 90°. So there should be 4 points.
How many points on the line x + y = 4 are there.............
Four points but two each on the two lines.
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by kstv » Wed Mar 17, 2010 8:19 am
This post is dying from unrequited answer. OA please?
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by firdaus117 » Wed Mar 17, 2010 10:02 am
Only two points viz p1 and p2 are possible on number line x+y=4 that lie at a distance of 1 unit from other line 4 x + 3 y = 10.

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by harshavardhanc » Wed Mar 17, 2010 11:21 am
sanju09 wrote:How many points on the line x + y = 4 are there that lie at a unit distance from the line 4 x + 3 y = 10?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
IMO, there will be only two points, lying on either side of the intersection.
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by treker » Wed Mar 17, 2010 7:50 pm
Can someone explain the logic behind solving this question?

Thanks!
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by kstv » Wed Mar 17, 2010 8:32 pm
But I think some restriction in the Q is needed. Otherwise why not the blue lines in the figure ?
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by [email protected] » Wed Mar 17, 2010 9:26 pm
treker wrote:Can someone explain the logic behind solving this question?

Thanks!
If you solve for the slopes of both lines

line 1. x + y = 4
y=-x+4
slope = -1

line 2. 4x + 3y = 10
y=(-4/3)x+10/3
slope = -4/3

Since they are not parallel, they are bound to intersect at some point. If you look at near the point of intersection, there can be only 2 points on line 1 that are at unit distance from line 2(on either sides of line 2).

If the lines are parallel and the distance between the lines is 1, there will be infinite number of points that are at unit distance. If the line is parallel and the distance betweeen the lines in not 1, there will be 0 points that are at unit distance.
Hope this helps ?!
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by harshavardhanc » Wed Mar 17, 2010 11:02 pm
kstv wrote:But I think some restriction in the Q is needed. Otherwise why not the blue lines in the figure ?
kstv,

whenever we say that a point lies at certain distance from a line, we talk about the perpendicular distance from that point to the line.

So, there will be only two points on the line : x+ y = 4, from where you can drop perpendiculars on the second line, because they intersect.

Now, as far as the blue lines in the figure are concerned, yes, they can be perpendiculars dropped from two points on the SECOND line on the first. But, we are not asked about them. The question says, points on first.

If you try to draw a figure, it will be clearer.
Regards,
Harsha
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by kstv » Thu Mar 18, 2010 6:18 am
harshavardhanc wrote:
kstv wrote:But I think some restriction in the Q is needed. Otherwise why not the blue lines in the figure ?
kstv,
whenever we say that a point lies at certain distance from a line, we talk about the perpendicular distance from that point to the line.....................If you try to draw a figure, it will be clearer.
Yes, just a small correction. It is perdendicular distance when we talk of the shortest distance. Here the question does not put that restriction it just says 1 unit. The Q can be made made more fool proof .
Firdaus is right cos' as per my logic there will be innumerable points on x+y=4 but the options are 1 - 5 so 2 will do. Thanks for evaluating diff facets of the problem.
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by harshavardhanc » Thu Mar 18, 2010 6:23 am
kstv wrote:
harshavardhanc wrote:
kstv wrote:But I think some restriction in the Q is needed. Otherwise why not the blue lines in the figure ?
kstv,
whenever we say that a point lies at certain distance from a line, we talk about the perpendicular distance from that point to the line.....................If you try to draw a figure, it will be clearer.
Yes, just a small correction. It is perdendicular distance when we talk of the shortest distance. Here the question does not put that restriction it just says 1 unit. The Q can be made made more fool proof .
Firdaus is right cos' as per my logic there will be innumerable points on x+y=4 but the options are 1 - 5 so 2 will do. Thanks for evaluating diff facets of the problem.
yes, i like this discussion too! thanks to you as well :)
Regards,
Harsha
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