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You can not rely on the graphical solution. I mean you can plot it, but then eventually you will have to play with the equation for distance between two points. _________________

Thanks Gentlemen for the algebraic approach i was using the graphical approach and was confused

adishail wrote:

A.

You can not rely on the graphical solution. I mean you can plot it, but then eventually you will have to play with the equation for distance between two points.

Actually you can solve this graphically. What you need to note is that points (m,n) and (n,m) are mirror reflections across the x=y (45 degree) line on the coordinate plane.

And the final piece in the puzzle is, if you draw a horizontal and a vertical line from (m,n) & the point (n,m) to intersect x=y, both lines will intersect at the points (m,m) and (n,n). This square you form has side (m-n) & its diagnol is the straight line joining (m,n) and (n,m)

So distance = Sqrt(2) * (m-n)

Hence all you need is m-n

The solution seems a bit involved, but if you draw it out it'll be pretty straight forward. _________________

Thanks Gentlemen for the algebraic approach i was using the graphical approach and was confused

adishail wrote:

A.

You can not rely on the graphical solution. I mean you can plot it, but then eventually you will have to play with the equation for distance between two points.

Actually you can solve this graphically. What you need to note is that points (m,n) and (n,m) are mirror reflections across the x=y (45 degree) line on the coordinate plane.

And the final piece in the puzzle is, if you draw a horizontal and a vertical line from (m,n) & the point (n,m) to intersect x=y, both lines will intersect at the points (m,m) and (n,n). This square you form has side (m-n) & its diagnol is the straight line joining (m,n) and (n,m)

So distance = Sqrt(2) * (m-n)

Hence all you need is m-n

The solution seems a bit involved, but if you draw it out it'll be pretty straight forward.

Actually, it is the same thing as plotting two points and then writing the equation of distance between two points. As far as being a faster way, the fact that the pyth theorem uses a difference is enough to get the soultion. _________________

Ultimately it has to be the same thing, same problem same solution ... just saying that you can solve the question by plotting a graph and without knowing the distance formula

Ultimately it has to be the same thing, same problem same solution ... just saying that you can solve the question by plotting a graph and without knowing the distance formula

Re: Co-ordinate geometry DS [#permalink]
09 May 2013, 20:46

Expert's post

josemnz83 wrote:

I'm not understanding how A is sufficient to answer the question. All A says is that m-n=2. What about n-m? How are m-n and n-m the same?

gurpreetsingh wrote:

Distance between two points (x_1,y_1) and (x_2,y_2) is \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

put the values you will get \sqrt{(m - n)^2 + (n - m)^2}

= \sqrt{2(m - n)^2}

Hence A is sufficient.

(m-n) and (n-m) are not the same thing, however, (m-n)^2 and (n-m)^2 are the same thing. Take (m-n) = 2, thus (n-m) = -2. However, (m-n)^2 = (n-m)^2 = 4. _________________