Question

The distance between two points (1, 1) and $ \left( {\dfrac{{2{t^2}}}{{1 + {t^2}}},\dfrac{{{{(1 - t)}^2}}}{{1 + {t^2}}}} \right) $ is
A. $ 4t $
B. $ 3t $
C. 1
D. None of these

Answer 1

Hint: To find out the distance between two coordinates we will apply the formula mentioned below and put the given coordinates into the formula, then simplify the equation with the help of L.C.M. and identities.
Formula used:
The distance between points $ ({x_1},{x_2}) $ and $ ({y_1},{y_2}) $ is: $ \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Identity: $ {(a - b)^2} = {a^2} + {b^2} - 2ab $

Complete step-by-step answer:
Two coordinates (1, 1) and $ \left( {\dfrac{{2{t^2}}}{{1 + {t^2}}},\dfrac{{{{(1 - t)}^2}}}{{1 + {t^2}}}} \right) $ are given in the question, to find the distance between them we use formula mentioned above.
Let the distance between two points: $ d $
 $ d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ (This is the formula we have mentioned above)
We will put the given values of coordinates in the above equation
\[d = \sqrt {{{\left( {\dfrac{{2{t^2}}}{{1 + {t^2}}} - 1} \right)}^2} + {{\left( {\dfrac{{{{(1 - t)}^2}}}{{1 + {t^2}}} - 1} \right)}^2}} \]
We will take L.C.M. of both denominators separately
 $ = \sqrt {{{\left( {\dfrac{{2{t^2} - (1 + {t^2})}}{{1 + {t^2}}}} \right)}^2} + {{\left( {\dfrac{{{{(1 - t)}^2} - (1 + {t^2})}}{{1 + {t^2}}}} \right)}^2}} $
Now, we will simplify the above equation by opening brackets and apply an identity $ {(a - b)^2} = {a^2} + {b^2} - 2ab $ on the numerator of second term
\[ = \sqrt {{{\left( {\dfrac{{2{t^2} - 1 - {t^2}}}{{1 + {t^2}}}} \right)}^2} + {{\left( {\dfrac{{{1} + {{{t}}^2} - 2t - {1} - {{{t}}^2}}}{{1 + {t^2}}}} \right)}^2}} \Rightarrow \sqrt {{{\left( {\dfrac{{{t^2} - 1}}{{1 + {t^2}}}} \right)}^2} + {{\left( {\dfrac{{ - 2t}}{{1 + {t^2}}}} \right)}^2}} \] (Similar terms with opposite signs will cancel out)
Again we apply an identity $ {(a - b)^2} = {a^2} + {b^2} - 2ab $ and expand terms
\[ = \sqrt {\dfrac{{{{\left( {{t^2} - 1} \right)}^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}} + \dfrac{{{{( - 2t)}^2}}}{{{{(1 + {t^2})}^2}}}} \Rightarrow \sqrt {\dfrac{{\left( {{t^4} + 1 - 2{t^2}} \right)}}{{\left( {1 + {t^4} + 2{t^2}} \right)}} + \dfrac{{(4{t^2})}}{{\left( {1 + {t^4} + 2{t^2}} \right)}}} \]
As the denominator is the same it’s easy to let the denominator as L.C.M., as we are letting in below step
\[ = \sqrt {\dfrac{{{t^4} + 1 - 2{t^2} + 4{t^2}}}{{1 + {t^4} + 2{t^2}}}} \]
Simplify the above equation by performing a subtraction operation
\[ = \sqrt {\dfrac{{{t^4} + 1 + 2{t^2}}}{{1 + {t^4} + 2{t^2}}}} \Rightarrow 1\]
Because the numerator and denominator are the same as shown in the above equation, therefore we will cancel out the whole numerator with denominator and the answer will be 1.
Hence, the distance between points (1, 1) and $ \left( {\dfrac{{2{t^2}}}{{1 + {t^2}}},\dfrac{{{{(1 - t)}^2}}}{{1 + {t^2}}}} \right) $ is 1unit. From the above options c is the right option.
So, the correct answer is “Option C”.

Note: We need to take care that, we should use identities while simplification to expand the equation and also to compact the equation as we used identity $ {(a - b)^2} = {a^2} + {b^2} - 2ab $ to get the endpoint easily.