Question

Find the distance between the points (0, 0) and (36, 15).

Answer 1

Hint:At first, name the point (0, 0) as O and (36, 15) as P, then draw a perpendicular from P to Q such that it's coordinates are (36,0). Then, finally apply Pythagoras theorem which says that, $O{{P}^{2}}=O{{Q}^{2}}+P{{Q}^{2}}$ to find the value of OP.

Complete step by step answer:
In the question, we are given two points (0, 0) and (36, 15) and we have to find the distance between them.
So, let’s consider (0, 0) point as O and (36, 15) as P. Hence, in other words, we have to find the length of OP.
So, at first, we can point it as:

Hence, Q is perpendicular drawn from P to x-axis so we can say its coordinates are (36, 0).
Thus, we can say that OQ is 36 units and PQ is 15 units.
Now to find OP we will use the Pythagoras theorem.
According to the Pythagoras theorem,
\[O{{P}^{2}}=O{{Q}^{2}}+P{{Q}^{2}}\]
We know OQ is 36 and PQ is 15 so on substituting we get,
\[O{{P}^{2}}={{36}^{2}}+{{15}^{2}}\]
On simplification we get:
\[\begin{align}
  & O{{P}^{2}}=1296+225 \\
 & O{{P}^{2}}=1521 \\
\end{align}\]
So the value of OP is $\sqrt{1521}\Rightarrow 39$
Hence, the distance between two points is 39.
Note:
Students can also do this same problem by using the distance formula which is, \[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] , where the coordinate of the points are $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ . So, here we have $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ as (0, 0) and (36, 15). So, if we apply the distance formula, we will end up with the same result as above, i.e. $\sqrt{1521}\Rightarrow 39$ .