lie at a unit

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by sanju09 » Fri Mar 19, 2010 1:22 am
Length of perpendicular or distance of a point (p, q) from a line a x + b y + c = 0 is

∣a p + b q + c∣ / √(p^2+ q^2)

If (k, 4 - k) is a point on the line x + y = 4 that lie at a unit distance from the line 4 x + 3 y = 10, then

∣4 k + 12 - 3 k - 10∣ / √[4^2 + (4 - k)^2] = 1

∣k + 2∣ = √[k^2 - 8 k + 32]

squaring both sides and solving

k = 7/3.

Hence, the unique point must be (7/3, 5/3).

My protest! If there were two such points on the line, we should have got two distinct values of k here.

How about [spoiler]A[/spoiler]?
The mind is everything. What you think you become. -Lord Buddha



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by harshavardhanc » Fri Mar 19, 2010 2:08 am
sanju09 wrote:
∣4 k + 12 - 3 k - 10∣ / √[4^2 + (4 - k)^2] = 1

Check the denominator once again.

Shouldn't it be K^2 + (4 - K )^2 ?

You will get 2 values. :)
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Harsha

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by sanju09 » Fri Mar 19, 2010 3:14 am
harshavardhanc wrote:
sanju09 wrote:
∣4 k + 12 - 3 k - 10∣ / √[4^2 + (4 - k)^2] = 1

Check the denominator once again.

Shouldn't it be K^2 + (4 - K )^2 ?

You will get 2 values. :)
Are you sure that it's without a square root sign? Check it once!
The mind is everything. What you think you become. -Lord Buddha



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by harshavardhanc » Fri Mar 19, 2010 3:31 am
sanju09 wrote:
harshavardhanc wrote:
sanju09 wrote:
∣4 k + 12 - 3 k - 10∣ / √[4^2 + (4 - k)^2] = 1

Check the denominator once again.

Shouldn't it be K^2 + (4 - K )^2 ?

You will get 2 values. :)
Are you sure that it's without a square root sign? Check it once!
Buddy, it's with the square root. But, that's not what I wanted to point out.

The point that I wanted you to notice was that in your original equation/formula, you haven mistakenly taken incorrect values for p & q.

Whereas, p should be K, you have taken it as 4. That's why in solving, you didn't get a quadratic equation.
Regards,
Harsha

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by sanju09 » Fri Mar 19, 2010 3:36 am
harshavardhanc wrote:
sanju09 wrote:
harshavardhanc wrote:
sanju09 wrote:
∣4 k + 12 - 3 k - 10∣ / √[4^2 + (4 - k)^2] = 1

Check the denominator once again.

Shouldn't it be K^2 + (4 - K )^2 ?

You will get 2 values. :)
Are you sure that it's without a square root sign? Check it once!
Buddy, it's with the square root. But, that's not what I wanted to point out.

The point that I wanted you to notice was that in your original equation/formula, you haven mistakenly taken incorrect values for p & q.

Whereas, p should be K, you have taken it as 4. That's why in solving, you didn't get a quadratic equation.
oh yeah

now you can rock, because the OA is [spoiler]B[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com