Find a relationship between x and y so that the triangle whose vertices are given by (x, y), (1,1) and (5, 1) is a right triangle with the hypotenuse defined by the points (1, 1) and (5, 1).
A. ${\left( {x + 3} \right)^2} - {\left( {y - 1} \right)^2} = {2^2}$
B. ${\left( {x - 3} \right)^2} + {\left( {y - 1} \right)^2} = {2^2}$
C. ${x^2} + {y^2} = {2^2}$
D. ${\left( {x - 3} \right)^2} - {\left( {y - 2} \right)^2} = {3^2}$

Hint: We will solve this problem by using the distance formula;
Formula: $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ .


Complete step by step solution:
Let us use the distance formula to find the length of the hypotenuse h with points (1, 1) and (5, 1).
$h = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ … (i)
Substitute the value of ${x_1},\,{y_1},{x_2}$ and ${y_2}$in equation (i).
$h = \sqrt {{{\left( {5 - 1} \right)}^2} + {{\left( {1 - 1} \right)}^2}} $
$h = \sqrt {{{\left( 4 \right)}^2} + {{\left( 0 \right)}^2}} $
$h = \sqrt {16 + 0} \Rightarrow \sqrt {16} $
$h = 4$
Hypotenuse is$4$.
Now we use the distance formula to find the sizes of the two other sides a and b of the triangle.
$a = \sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 1} \right)}^2}} $
$b = \sqrt {{{\left( {x - 5} \right)}^2} + {{\left( {y - 1} \right)}^2}} $
By Pythagoras theorem gives
${4^2} = {\left( {\sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 1} \right)}^2}} } \right)^2} + {\left( {\sqrt {{{\left( {x - 5} \right)}^2} + {{\left( {y - 1} \right)}^2}} } \right)^2}$
Expand the squares, simplify and complete the square to rewrite above relationship between${\left( {x - 3} \right)^2} + {\left( {y - 1} \right)^2} = {2^2}$.
${4^2} = \left( {{{\left( {x - 1} \right)}^2} + {{\left( {y - 1} \right)}^2}} \right) + \left( {{{\left( {x - 5} \right)}^2} + {{\left( {y - 1} \right)}^2}} \right)$
$ = \left( {{x^2} + 1 - 2x + {y^2} + 1 - 2y} \right) + \left( {{x^2} + 25 - 10x + {y^2} + 1 - 2y} \right)$
$ = {x^2} + 1 - 2x + {y^2} + 1 - 2y + {x^2} + 25 - 10x + {y^2} + 1 - 2y$
$ = 2{x^2} + 2{y^2} - 12x - 4y + 28$
${4^2} = 2\left( {{x^2} + {y^2} - 6x - 2y + 14} \right)$
$\dfrac{{16}}{2} = ({x^2} - 6x + 9) + ({y^2} - 2y + 1) + 4$
$8 - 4 = {(x - 3)^2} + {(y - 1)^2}$
${a^2} - 2ab + {b^2} = {(a - b)^2}$
${(1.1)^{10000}}$
Then, we will break the value ${(1.1)^{10000}}$into two parts such that
$
  {(1 + 0.1)^{10000}} \\
  {(1 + 0.1)^{10000}} = {(1 + 0.1)^{10000}} \\
 $
We know that binomial theorem,
\[{(x + a)^x} = {\,^n}{C_0}{a^x} + {\,^n}{C_1}{a^{x - 1}}{x^1} + {\,^n}{C_2}{a^{x - 2}}{x^2} + ...... + {\,^n}{C_x}{x^2}\]
So, $x = 1,\,\,a = 0.1\,\,or\,\,\,n = 10000$
Then will put these values in the above binomial theorem.
Then
\[{(1 + 0.1)^{10000}} = {\,^{10000}}{C_0}{(1)^{10000}} + {\,^{10000}}{C_1}{(1)^{10000}} \times (0.1) + ......{(1 + 0.1)^{10000}} = \dfrac{{10000!}}{{0!10000 - 0!}} \times {(1)^{10000}} + \dfrac{{10000 \times 9999!}}{{1! \times 9999!}}{(1)^{9999}} \times (0.1)\]
${(1 + 0.1)^{10000}} = \dfrac{{100001}}{{100001}} + 10000 \times 0.1$
$
  {(1 + 0.1)^{0000}} = 1 + 1000.0 \\
  {(1 + 0.1)^{10000}} = 1001 \\
 $
So, ${(1 + 0.1)^{10000}} > 1000$


Note: Students must keep in mind that Pythagoras theorem will apply only for the right angle triangle and it will give $h{(hyp.)^2} = {(base)^2} + {(perp.)^2}$.