Question

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

Answer 1

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 $\left( -2,-3 \right)$.

& {{x}_{1}}=-2......(1) \\\ & {{y}_{1}}=-3.......(2) \\\ \end{aligned}$$ Now let us compare $$B\left( {{x}_{2}},{{y}_{2}} \right)$$ with $$\left( 3,2 \right)$$. $$\begin{aligned} & {{x}_{2}}=3......(3) \\\ & {{y}_{2}}=2.......(4) \\\ \end{aligned}$$ We know that the distance between $$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{aligned} & \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{aligned}$$ From equation (1), it is clear that the value of d is equal to $$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{aligned} & {{x}_{1}}=-2......(1) \\\ & {{y}_{1}}=-3.......(2) \\\ \end{aligned}$$ Now let us compare $$B\left( {{x}_{2}},{{y}_{2}} \right)$$ with $$\left( 3,2 \right)$$. $$\begin{aligned} & {{x}_{2}}=3......(3) \\\ & {{y}_{2}}=2.......(4) \\\ \end{aligned}$$ Let us assume the distance between $$\left( -2,-3 \right)\text{ and }\left( 3,2 \right)$$ is equal to d. $$\begin{aligned} & \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{aligned}$$ From equation (1), it is clear that the value of d is equal to $$\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.