Question

How do you find the distance between $(5,5)$ and $\,(4,4)$ ?

Answer 1

Hint: First we will mention all the coordinates and their respective variables accordingly. Then mention the distance formula which is given by \[d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \] and then substitute the values and hence evaluate the distance.

Complete step-by-step solution:
We will start off by mentioning all the coordinates and their respective variables accordingly.
$
  {({x_1},{y_1})} = {(5,5)} \\
  {({x_2},{y_2})} = {(4,4)} \\

$
The distance between two points is the length of the interval joining the two points.
For evaluating the distance between the two points we will use the distance formula.
Distance formula is given by,
\[d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
Now substitute the values in the distance formula and solve for the value of distance.
\[
\Rightarrow d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\
\Rightarrow d = \sqrt {{{(4 - 5)}^2} + {{(4 - 5)}^2}} \\
\Rightarrow d = \sqrt {{{( - 1)}^2} + {{( - 1)}^2}} \\
\Rightarrow d = \sqrt {1 + 1} \\
\Rightarrow d = \sqrt 2 \\
 \]
Hence, the distance between the points is \[\sqrt 2 \].

Additional information: The distance formula is a useful tool in finding the distance between two points which can be arbitrarily represented as points $({x_1},{y_1})$ and $({x_2},{y_2})$. The distance formula itself is actually derived from the Pythagoras Theorem which is ${a^2} + {b^2} = {c^2}$ where $c$ is the longest side of a right triangle which is also called as the hypotenuse and $a$ and $b$ are the other shorter sides which are known as the legs of the right triangle. The very essence of the Distance Formula is to calculate the length of the hypotenuse of the right triangle which is represented by the letter $c$.

Note: While solving for the distance between the two points, substitute the values in the distance formula along with their signs. While comparing values of terms with the general equation, compare along with their respective signs. Substitute values carefully in the formulas for evaluating values of terms.