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DS - XY plane

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by karthikpandian19 » Wed Jul 04, 2012 12:17 am
If line a and line b lie in the xy-plane, line a has a slope of (k - 1), and line b passes through the point (-2, k), are lines a and b perpendicular to one another?

Line a has a y-intercept of 8k - 6.
The two lines intersect at point (-6, 2k).
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by niketdoshi123 » Wed Jul 04, 2012 4:55 am
karthikpandian19 wrote:If line a and line b lie in the xy-plane, line a has a slope of (k - 1), and line b passes through the point (-2, k), are lines a and b perpendicular to one another?

Line a has a y-intercept of 8k - 6.
The two lines intersect at point (-6, 2k).
Slopes of two perpendicular lines are -ve reciprocals of each other.
given, slope of line a = (k-1)
and line b passes through the point (-2,k)
for lines a & b to be perpendicular to one another, the slope of line b (must be) = -1/(k-1)

Statement 1INSUFFICIENT
Doesn't give any information about the slope of line b

Statement 2 SUFFICIENT
Since both the lines intersect at point(-6,2k)
=> line b passes through point (-6,2k)
Now we have 2 points through which line b passes so we can find out the slope of line b.
Hence we can compare the the slopes of both the lines.

IMO the answer is B
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by eagleeye » Wed Jul 04, 2012 8:32 am
karthikpandian19 wrote:If line a and line b lie in the xy-plane, line a has a slope of (k - 1), and line b passes through the point (-2, k), are lines a and b perpendicular to one another?

Line a has a y-intercept of 8k - 6.
The two lines intersect at point (-6, 2k).
We are told that Line a has slope of k-1. And we need to find whether lines a and b are perpendicular. Recall that the product of slopes of two lines is -1 if they are perpendicular. Hence we need to find whether slope of line b is -1/(k-1). We are also told that line b passes through (-2,k).
Let the equation of line b be y=mx+b.
Then we have k=-2m+b.

With this in mind, let's look at the options.

1. Line a has a y-intercept of 8k - 6.
This doesn't tell us anything about line b. Insufficient.

2. The two lines intersect at point (-6, 2k).
This tells us something about b. We had from question stem, k=-2m+b.
If b passes through (-6,2k), we have 2k=-6m+b
Subtracting the two we get: k=-4m. so m=-k/4. This is not a multiple of -1/k-1. Hence depending on the value of k, The lines may or may not be perpendicular. Insufficient.
Also, as a passes through (-6,2k),
we get that for line a:
2k=-6(k-1) + a
=> a=8k-6. So y-intercept of a is 8k-6.

Combining the two, equation 1 doesn't add anything as we already figured out that y-intercept of a is 8k-6 from statement 2. Hence together, they are insufficient as well.

Hence the final answer is E.

Let me know if this helps :)
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by karthikpandian19 » Wed Jul 04, 2012 6:20 pm
Reasoning for statement 2 is wrong

(sorry i wrongly pressed the THANK button)
niketdoshi123 wrote:
karthikpandian19 wrote:If line a and line b lie in the xy-plane, line a has a slope of (k - 1), and line b passes through the point (-2, k), are lines a and b perpendicular to one another?

Line a has a y-intercept of 8k - 6.
The two lines intersect at point (-6, 2k).
Slopes of two perpendicular lines are -ve reciprocals of each other.
given, slope of line a = (k-1)
and line b passes through the point (-2,k)
for lines a & b to be perpendicular to one another, the slope of line b (must be) = -1/(k-1)

Statement 1INSUFFICIENT
Doesn't give any information about the slope of line b

Statement 2 SUFFICIENT
Since both the lines intersect at point(-6,2k)
=> line b passes through point (-6,2k)
Now we have 2 points through which line b passes so we can find out the slope of line b.
Hence we can compare the the slopes of both the lines.

IMO the answer is B
Regards,
Karthik
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by karthikpandian19 » Wed Jul 04, 2012 6:23 pm
I lost the track starting from "m=-k/4. This is not a multiple of -1/k-1. Hence depending on the value of k, The lines may or may not be perpendicular. Insufficient."

How do you determine it is not a multiple ????? Can you explain?

Anyway the OA is E

eagleeye wrote:
karthikpandian19 wrote:If line a and line b lie in the xy-plane, line a has a slope of (k - 1), and line b passes through the point (-2, k), are lines a and b perpendicular to one another?

Line a has a y-intercept of 8k - 6.
The two lines intersect at point (-6, 2k).
We are told that Line a has slope of k-1. And we need to find whether lines a and b are perpendicular. Recall that the product of slopes of two lines is -1 if they are perpendicular. Hence we need to find whether slope of line b is -1/(k-1). We are also told that line b passes through (-2,k).
Let the equation of line b be y=mx+b.
Then we have k=-2m+b.

With this in mind, let's look at the options.

1. Line a has a y-intercept of 8k - 6.
This doesn't tell us anything about line b. Insufficient.

2. The two lines intersect at point (-6, 2k).
This tells us something about b. We had from question stem, k=-2m+b.
If b passes through (-6,2k), we have 2k=-6m+b
Subtracting the two we get: k=-4m. so m=-k/4. This is not a multiple of -1/k-1. Hence depending on the value of k, The lines may or may not be perpendicular. Insufficient.
Also, as a passes through (-6,2k),
we get that for line a:
2k=-6(k-1) + a
=> a=8k-6. So y-intercept of a is 8k-6.

Combining the two, equation 1 doesn't add anything as we already figured out that y-intercept of a is 8k-6 from statement 2. Hence together, they are insufficient as well.

Hence the final answer is E.

Let me know if this helps :)
Regards,
Karthik
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by eagleeye » Wed Jul 04, 2012 6:59 pm
karthikpandian19 wrote:I lost the track starting from "m=-k/4. This is not a multiple of -1/k-1. Hence depending on the value of k, The lines may or may not be perpendicular. Insufficient."

How do you determine it is not a multiple ????? Can you explain?
Sure karthik:

I can see how that might be confusing. Let me rephrase. We know that slope of Line a is k-1. For us to get a sure answer of whether Line b with slope "m" is perpendicular, for all values of k, we need that:

m * (k-1) = r (where r is any number).
=> m = r/(k-1)
Here are a few examples. This is a unique question because if we get any number value of the product other than -1, we can certainly say that the lines are not perpendicular.

If m = 0, r = 0, so lines are not perpendicular.
If m = 1/(k-1), r = 1, the lines are not perpendicular.
If m = -1/(k-1), r = -1, the lines are perpendicular.
If m = 2.345/(k-1), r = 2.345, the lines are not perpendicular
If m = -pi/(k-1), r = -pi, the lines are not perpendicular.

So if we had gotten m in the form m = (Any number)/(k-1), we would be able to say sufficiently whether the lines are perpendicular. Since -k/4 is not in the form above, we can't say for sure.

To go further into this, (even though the question doesn't require this) , we would have the equation to check for orthogonality (perpendicular-ness), which would be (-k/4)*(k-1)=-1. This would give us two values for which Line a and Line b would be perpendicular. All other values of k would give us non-perpendicular lines.

Let me know if this helps :)
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by niketdoshi123 » Wed Jul 04, 2012 8:55 pm
eagleeye wrote:
karthikpandian19 wrote:I lost the track starting from "m=-k/4. This is not a multiple of -1/k-1. Hence depending on the value of k, The lines may or may not be perpendicular. Insufficient."

How do you determine it is not a multiple ????? Can you explain?
Sure karthik:

I can see how that might be confusing. Let me rephrase. We know that slope of Line a is k-1. For us to get a sure answer of whether Line b with slope "m" is perpendicular, for all values of k, we need that:

m * (k-1) = r (where r is any number).
=> m = r/(k-1)
Here are a few examples. This is a unique question because if we get any number value of the product other than -1, we can certainly say that the lines are not perpendicular.

If m = 0, r = 0, so lines are not perpendicular.
If m = 1/(k-1), r = 1, the lines are not perpendicular.
If m = -1/(k-1), r = -1, the lines are perpendicular.
If m = 2.345/(k-1), r = 2.345, the lines are not perpendicular
If m = -pi/(k-1), r = -pi, the lines are not perpendicular.

So if we had gotten m in the form m = (Any number)/(k-1), we would be able to say sufficiently whether the lines are perpendicular. Since -k/4 is not in the form above, we can't say for sure.

To go further into this, (even though the question doesn't require this) , we would have the equation to check for orthogonality (perpendicular-ness), which would be (-k/4)*(k-1)=-1. This would give us two values for which Line a and Line b would be perpendicular. All other values of k would give us non-perpendicular lines.

Let me know if this helps :)
While solving I got this equation k^2 - k - 4 = 0, but did not consider it since the roots were not integers. I guess this was the mistake as the question stem doesn't say k should be an integer.
Thanks for clarifying.:-)

Niket
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by karthikpandian19 » Wed Jul 04, 2012 11:01 pm
understood...thank you for your explanation
eagleeye wrote:
karthikpandian19 wrote:I lost the track starting from "m=-k/4. This is not a multiple of -1/k-1. Hence depending on the value of k, The lines may or may not be perpendicular. Insufficient."

How do you determine it is not a multiple ????? Can you explain?
Sure karthik:

I can see how that might be confusing. Let me rephrase. We know that slope of Line a is k-1. For us to get a sure answer of whether Line b with slope "m" is perpendicular, for all values of k, we need that:

m * (k-1) = r (where r is any number).
=> m = r/(k-1)
Here are a few examples. This is a unique question because if we get any number value of the product other than -1, we can certainly say that the lines are not perpendicular.

If m = 0, r = 0, so lines are not perpendicular.
If m = 1/(k-1), r = 1, the lines are not perpendicular.
If m = -1/(k-1), r = -1, the lines are perpendicular.
If m = 2.345/(k-1), r = 2.345, the lines are not perpendicular
If m = -pi/(k-1), r = -pi, the lines are not perpendicular.

So if we had gotten m in the form m = (Any number)/(k-1), we would be able to say sufficiently whether the lines are perpendicular. Since -k/4 is not in the form above, we can't say for sure.

To go further into this, (even though the question doesn't require this) , we would have the equation to check for orthogonality (perpendicular-ness), which would be (-k/4)*(k-1)=-1. This would give us two values for which Line a and Line b would be perpendicular. All other values of k would give us non-perpendicular lines.

Let me know if this helps :)
Regards,
Karthik
The source of the questions that i post from JUNE 2013 is from KNEWTON

---If you find my post useful, click "Thank" :) :)---
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