Question

The distance between the points (1,1) and (2t^2/1+t^2),(1-t)^2/1+t^2). Plz answer step by step.

Answer 1

GIVEN :

The points are (1,1) and $(\frac{2t^2}{1+t^2},\frac{(1-t)^2}{1+t^2})$

TO FIND :

The distance between the given two points.

SOLUTION :

Given that the two points are (1,1) and $(\frac{2t^2}{1+t^2},\frac{(1-t)^2}{1+t^2})$

Now we have to find the distance between the given two points :

Let the point (1,1) be $(x_1,y_1)$ and let the point $(\frac{2t^2}{1+t^2},\frac{(1-t)^2}{1+t^2})$ be $(x_2,y_2)$

The formula for distance between the two points is given by :

$s=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ units

By substituting  the values in the above distance formula we get,

$s=\sqrt{(\frac{2t^2}{1+t^2}-1)^2+(\frac{(1-t)^2}{1+t^2}-1)^2}$ units

$=\sqrt{(\frac{2t^2-1-t^2}{1+t^2})^2+(\frac{1^2+t^2-2t-1-t^2}{1+t^2})^2}$

$=\sqrt{(\frac{t^2-1}{1+t^2})^2+(\frac{-2t}{1+t^2})^2}$

By using the property of Exponents :

$(\frac{a}{b})^m=\frac{a^m}{b^m}$

$=\sqrt{\frac{(t^2-1)^2}{(1+t^2)^2}+\frac{(-2t)^2}{(1+t^2)^2}}$

By using the Algebraic identity :

$(a+b)^2=a^2+b^2+2ab$

$=\sqrt{\frac{t^4-2t^2+1^2}{(1+t^2)^2}+\frac{4t^2}{(1+t^2)^2}}$

$=\sqrt{\frac{t^4-2t^2+1+4t^2}{(1+t^2)^2}}$

$=\sqrt{\frac{t^4+1+2t^2}{(1+t^2)^2}}$

By using the Algebraic identity :

$(a+b)^2=a^2+b^2+2ab$

$=\sqrt{\frac{(t^2+1)^2}{(1+t^2)^2}}$

$=\sqrt{\frac{(1+t^2)^2}{(1+t^2)^2}}$

$=\sqrt{1}$

=1 unit

∴ s=1 unit.

∴ the distance between the given two points (1,1) and $(\frac{2t^2}{1+t^2},\frac{(1-t)^2}{1+t^2})$ is 1 unit.

Answer 2

Answer:

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