Question

Find the distance of the point $\left( 36,15 \right)$ from the origin.

Answer 1

Hint: In this question, we will use distance formula for finding the distance between two points in two-dimensional geometry formula, to find distance between given point and origin$\left( 0,0 \right)$.

Complete step-by-step answer:
Let $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$be two points in two-dimensional geometry. Then distance, let say $d$, between these two points is given as,
$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\cdots \cdots \left( i \right)$
In the given question, one point is $\left( 36,15 \right)$ and distance is to be measured from origin, that is the point $\left( 0,0 \right)$.
Here, from equation $\left( i \right)$, ${{x}_{1}}=36,\,{{y}_{1}}=15,\,{{x}_{2}}=0,\,{{y}_{2}}=0$.
Therefore, distance of $\left( 36,15 \right)$ from the origin, let say $d$, is given as below,
 $\begin{align}
  & d=\sqrt{{{\left( 0-36 \right)}^{2}}+{{\left( 0-15 \right)}^{2}}} \\
 & =\sqrt{{{\left( -36 \right)}^{2}}+{{\left( -15 \right)}^{2}}} \\
\end{align}$
Here, squaring negative signs, we will get positive signs, so we get,
$d=\sqrt{{{36}^{2}}+{{15}^{2}}}$
Here, writing prime factorisation of 36 and 15, which are 36 = ${{2}^{2}}\times {{3}^{2}}$ and 15 = $3\times 5$, we get,
$d=\sqrt{{{\left( {{2}^{2}}\times {{3}^{2}} \right)}^{2}}+{{\left( 3\times 5 \right)}^{2}}}$
Here, distributing power 2 inside brackets, we get,
$\begin{align}
  & d=\sqrt{\left( {{2}^{2\times 2}}\times {{3}^{2\times 2}} \right)+\left( {{3}^{2}}\times {{5}^{2}} \right)} \\
 & =\sqrt{\left( {{2}^{4}}\times {{3}^{4}} \right)+\left( {{3}^{2}}\times {{5}^{2}} \right)} \\
\end{align}$
Here, taking ${{3}^{2}}$ common inside square root sign, we get,
$d=\sqrt{{{3}^{2}}\left( {{2}^{4}}\times {{3}^{2}}+{{5}^{2}} \right)}$
Here, writing value of square root of ${{3}^{2}}$, we get,
$d=3\sqrt{{{2}^{4}}\times {{3}^{2}}+{{5}^{2}}}$
Here, writing the values of powers of the terms inside square root sign, we get,
$\begin{align}
  & d=3\sqrt{16\times 9+25} \\
 & =3\sqrt{144+25} \\
 & =3\sqrt{169} \\
\end{align}$
Here, we have, ${{13}^{2}}=169$, putting this value here, we get,
$d=3\sqrt{{{13}^{2}}}$
Here, writing value of square root of ${{13}^{2}}$, we get,
$\begin{align}
  & d=3\times 13 \\
 & =39 \\
\end{align}$
Hence, we get that the distance of the point $\left( 36,15 \right)$ from the origin is 39 units.

Note: In this type of question, when distance is asked from origin, you can directly take the root of sum of square of $x$ and $y$ coordinate of the given point to find the distance as we know that if we add and subtract 0 with any number, then the number does not change.