Question

Find the distance between the following pair of points \[\left( -2,-3 \right)\text{ and }\left( 3,2 \right)\]

Answer 1

Hint: We know that the distance between \[A\left( {{x}_{1}},{{y}_{1}} \right)\] and \[B\left( {{x}_{2}},{{y}_{2}} \right)\] is equal to AB if \[AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]. Now let us compare \[A\left( {{x}_{1}},{{y}_{1}} \right)\] with \[\left( -2,-3 \right)\]. Now let us compare \[B\left( {{x}_{2}},{{y}_{2}} \right)\] with \[\left( 3,2 \right)\]. Now by using the above formula, we can find the distance between the pair of points \[\left( -2,-3 \right)\text{ and }\left( 3,2 \right)\].

Complete step by step answer:
Before solving the problem, we should know that the distance between \[A\left( {{x}_{1}},{{y}_{1}} \right)\] and \[B\left( {{x}_{2}},{{y}_{2}} \right)\] is equal to AB if \[AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\].
From the question, it is clear that we should find the distance between the pair of points \[\left( -2,-3 \right)\text{ and }\left( 3,2 \right)\].

Now let us compare \[A\left( {{x}_{1}},{{y}_{1}} \right)\] with \[\left( -2,-3 \right)\].
\[\begin{align}
  & {{x}_{1}}=-2......(1) \\
 & {{y}_{1}}=-3.......(2) \\
\end{align}\]
Now let us compare \[B\left( {{x}_{2}},{{y}_{2}} \right)\] with \[\left( 3,2 \right)\].
\[\begin{align}
  & {{x}_{2}}=3......(3) \\
 & {{y}_{2}}=2.......(4) \\
\end{align}\]
We know that the distance between \[A\left( {{x}_{1}},{{y}_{1}} \right)\] and \[B\left( {{x}_{2}},{{y}_{2}} \right)\] is equal to AB if \[AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\].
Let us assume the distance between \[\left( -2,-3 \right)\text{ and }\left( 3,2 \right)\] is equal to d.
\[\begin{align}
  & \Rightarrow d=\sqrt{{{\left( 3-(-2) \right)}^{2}}+{{\left( 2-(-3) \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{{{\left( 3+2 \right)}^{2}}+{{\left( 2+3 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{{{\left( 5 \right)}^{2}}+{{\left( 5 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{25+25} \\
 & \Rightarrow d=\sqrt{\left( 25 \right)\left( 2 \right)} \\
 & \Rightarrow d=5\sqrt{2}.....(1) \\
\end{align}\]
From equation (1), it is clear that the value of d is equal to \[5\sqrt{2}\].

So, we can say that the distance between the following pair of points \[\left( -2,-3 \right)\text{ and }\left( 3,2 \right)\] is equal to \[5\sqrt{2}\].

Note: Students may have a misconception that that the distance between \[A\left( {{x}_{1}},{{y}_{1}} \right)\] and \[B\left( {{x}_{2}},{{y}_{2}} \right)\] is equal to AB if \[AB=\sqrt{{{\left( {{x}_{2}}+{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}+{{y}_{1}} \right)}^{2}}}\]. If this misconception is followed, then we get
From the question, it is clear that we should find the distance between the pair of points \[\left( -2,-3 \right)\text{ and }\left( 3,2 \right)\].
Now let us compare \[A\left( {{x}_{1}},{{y}_{1}} \right)\] with \[\left( -2,-3 \right)\].
\[\begin{align}
  & {{x}_{1}}=-2......(1) \\
 & {{y}_{1}}=-3.......(2) \\
\end{align}\]
Now let us compare \[B\left( {{x}_{2}},{{y}_{2}} \right)\] with \[\left( 3,2 \right)\].
\[\begin{align}
  & {{x}_{2}}=3......(3) \\
 & {{y}_{2}}=2.......(4) \\
\end{align}\]
Let us assume the distance between \[\left( -2,-3 \right)\text{ and }\left( 3,2 \right)\] is equal to d.
\[\begin{align}
  & \Rightarrow d=\sqrt{{{\left( 3+(-2) \right)}^{2}}+{{\left( 2+(-3) \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{{{\left( 3-2 \right)}^{2}}+{{\left( 2-3 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{1+1} \\
 & \Rightarrow d=\sqrt{2}.....(1) \\
\end{align}\]
From equation (1), it is clear that the value of d is equal to \[\sqrt{2}\].
So, we can say that the distance between the following pair of points \[\left( -2,-3 \right)\text{ and }\left( 3,2 \right)\] is equal to \[\sqrt{2}\]. While solving this problem, students should also avoid calculation mistakes. Because a single mistake will interrupt the final answer.