Question

If the ordered pairs (x, y) of positive integers x and y satisfying the equation tan–1x + cot–1y = tan–13 are represented as points A and B in cartesian plane, then distance between points A and B is

Answer 1

SOLUTION

TO DETERMINE

If the ordered pairs (x, y) of positive integers x and y satisfying the equation

$\displaystyle \sf{ { \tan}^{ - 1}x +{ \cot}^{ - 1} y = { \tan}^{ - 1}3 }$

are represented as points A and B in cartesian plane, then distance between points A and B is

EVALUATION

Here the given equation is

$\displaystyle \sf{ { \tan}^{ - 1}x +{ \cot}^{ - 1} y = { \tan}^{ - 1}3 }$

We solve it as below

$\displaystyle \sf{ { \tan}^{ - 1}x +{ \cot}^{ - 1} y = { \tan}^{ - 1}3 }$

$\displaystyle \sf{ \implies \: { \tan}^{ - 1}x +{ \tan}^{ - 1} \frac{1}{y} = { \tan}^{ - 1}3 }$

$\displaystyle \sf{ \implies \: { \tan}^{ - 1} \: \frac{x + \frac{1}{y} }{1 - \frac{x}{y} } = { \tan}^{ - 1}3 }$

$\displaystyle \sf{ \implies \: { \tan}^{ - 1} \: \frac{x y + 1}{y - x} = { \tan}^{ - 1}3 }$

$\displaystyle \sf{ \implies \: \: \frac{x y + 1}{y - x} =3 }$

$\displaystyle \sf{ \implies \: \: xy + 1 = 3y - 3x }$

$\displaystyle \sf{ \implies \: \: xy + 3x = 3y - 1 }$

$\displaystyle \sf{ \implies \: \: 3y - xy = 3x + 1 }$

$\displaystyle \sf{ \implies \: \: y = \frac{3x + 1}{3 - x} }$

Since x and y are positive integers

So x = 1 gives y = 2

x = 2 gives y = 7

So two pair of values of x and y in the form (x, y) represents the points A(1,2) , B(2,7) in cartesian plane

So the required required distance

= AB

$\sf{ = \sqrt{ {(2 - 1)}^{2} + {(7 - 2)}^{2} } }$

$\sf{ = \sqrt{ {(1)}^{2} + {(5)}^{2} } }$

$\sf{ = \sqrt{ 1 + 25 } }$

$\sf{ = \sqrt{ 26} }$

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