Question

Find the distance between the points \[(0,0)\] and \[(36,15)\]

Answer 1

Hint: We use the formula of distance between two points and calculate the distance by substituting the values of points in the formula. Find the value under the square root and try to write that value in terms of square of a number so as to cancel square root by square power. Use the method of prime factorization to write the number under the square root in simpler form.
* Distance between the two points \[(x,y)\] and \[(a,b)\] is given by\[D = \sqrt {{{(x - a)}^2} + {{(y - b)}^2}} \]where L can never be negative.
* Prime factorization is a process of writing a number in multiple of its factors where all factors are prime numbers.

Complete step-by-step solution:
We are given two points \[(0,0)\] and \[(36,15)\]
Compare the given points \[(0,0)\] and \[(36,15)\] to the general points \[(x,y)\] and \[(a,b)\].
On comparison we get the value of \[x = 0,y = 0,a = 36,b = 15\]
Since we know the formula of distance between two points \[(x,y)\] and \[(a,b)\] is given by \[D = \sqrt {{{(x - a)}^2} + {{(y - b)}^2}} \]
Substitute the values of \[x = 0,y = 0,a = 36,b = 15\] in the formula of distance
\[ \Rightarrow D = \sqrt {{{(0 - 36)}^2} + {{(0 - 15)}^2}} \]
Calculate the squares of numbers inside the brackets
\[ \Rightarrow D = \sqrt {1296 + 225} \]
Add the terms under the square root
\[ \Rightarrow D = \sqrt {1521} \]
Now we write the prime factorization of the number under the square root i.e. 1521
\[ \Rightarrow \]Prime factorization of \[1521 = 3 \times 3 \times 13 \times 13\]
We can add the powers when the base is same
\[ \Rightarrow \]Prime factorization of \[1521 = {3^2} \times {13^2}\]
Since powers are same the base can be multiplied
\[ \Rightarrow \]Prime factorization of \[1521 = {\left( {3 \times 13} \right)^2}\]
\[ \Rightarrow \]Prime factorization of \[1521 = {39^2}\]
Substitute the value of \[1521 = {39^2}\] under the square root in RHS of the equation
\[ \Rightarrow D = \sqrt {{{39}^2}} \]
Cancel square root by square power on right hand side of the equation
\[ \Rightarrow D = 39\]

\[\therefore \]Distance between the two points \[(0,0)\] and \[(36,15)\] is 39 units.

Note: Students are likely to make the mistake of writing the value of distance directly as \[\sqrt {1521} \] as they don’t know whose square value is 1521. Keep in mind we use a method of prime factorization to break the number into smaller parts and then combine the factors to form square terms.