Question

@ganeshie8 @mathslover

Answer 1

A family of lines is given by \((1 + 2\lambda) x + (1 - \lambda) y + \lambda = 0 \), where \(\lambda\) is the parameter.

The line belonging to this family at the maximum distance from the point \((1,4)\) is?

Answer 2

what is the question anonymous person?

Answer 3

Here is what I did:\[(x + y) + \lambda (2x - y + 1) = 0\]So the point of concurrency is \((1/3, -1/3)\). Now I wrote the line as\[y + 1/3 = m(x - 1/3)\]

Answer 4

Then I calculated the distance in terms of \(m\). But I have no idea how one maximizes or minimizes that.

Answer 5

Oh, whoops. The point of concurrency is \((-1/3,1/3)\)

Answer 6

\[y + 1/3 = m(x - 1/3)\]\[y - mx + \frac{m+1}{3} = 0\]

Answer 7

\[d = \dfrac{|4 - m + (m+1)/3|}{\sqrt{m^2 + 1}}\]

Answer 8

...aaaand I'm stuck. Any other ways to do this? Maybe keeping things in terms of lambda?

Answer 9

where's mathslover when I need him

Answer 10

considering different intervals and differentiating may help...

Answer 11

basically you're given a family of lines all passing through a fixed point and asked to find the maximum distance from a point

your method works perfectly!

Answer 12

Just that I don't know how to maximise/minimise. :|

Answer 13

OK, let's try it interval-by-interval.

Answer 14

\[4 - m + (m+1)/3 \ge 0 \]\[12 - 3m + m + 1 \ge 0\]\[2m \le 13 \]\[m \le 13/2\]So that's the case one, and our expression is given by\[\dfrac{\frac{m+1}{3} + 4 - m}{\sqrt{m^2 + 1}}\]

Answer 15

i think we can try maximizing square of distance to avoid the cases

Answer 16

ah, of course. why didn't I think of that?

Answer 17

\[\dfrac{m+1}{3} - m + 4 = \dfrac{-2m}{3} + \dfrac{13}{3}\]Squaring,\[\dfrac{4}{9}m^2 - \dfrac{52}{9} m + \frac{169}{9}\]Because we're maximizing, we can remove the fractions and stuff.\[f(m)= \frac{4m^2 - 52m + 169}{m^2 + 1}\]is what we need to maximize.

Answer 18

So this is a good-old rational expression...

... which, unfortunately, I'm too lazy to work with. WolframAlpha overlords!

Answer 19

http://www.wolframalpha.com/input/?i=maximize+%5Cdfrac%7B%284+-+m+%2B+%28m%2B1%29%2F3%29%5E2%7D%7Bm%5E2+%2B+1%7D

Answer 20

m = -2/13. Yup.

Answer 21

Hey Parth, sorry, was busy in some other work. Didn't notice that.
I am looking at your work now. Give me 2 minutes, sorry for the delay yaar!

Answer 22

\[y +1/3 = -2/13(x -1/3)\]

Answer 23

brb, battery low

Answer 24

i wonder how you get \(\lambda \) out of \(m\) hmm

Answer 25

comparing coefficients?

Answer 26

\[13y + 13/3 = -2x + 2/3\]\[2x + 13y + 11/3 = 0\]\[6x + 39y + 11 = 0\]which is of course the incorrect answer. :)