Question

Find the distance between the points A and B in the following:
(i) \[A\left( a+b,\text{ }b-a \right),B\left( a-b,\text{ }a+b \right)\]
(ii) \[A\left( 1,-1 \right),B\left( \dfrac{-1}{2},\dfrac{1}{2} \right)\]
(iii) \[A\left( 8,-2 \right),B\left( 3,-6 \right)\]
(iv) \[A\left( a+b,a-b \right),B\left( a-b,-a-b \right)\]

Answer 1

Hint: We will use the distance formula to find the distance between two points if we have two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] the distance formula is given by:
\[D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

Complete step-by-step answer:
So by using this formula we can calculate the distance between two points.

We know that the distance between any two points, say \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is given by:
\[D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

So we will use the above formula in each question shown as below:

(i) \[A\left( a+b,\text{ }b-a \right),B\left( a-b,\text{ }a+b \right)\]
We have \[{{x}_{1}}=a+b,{{y}_{1}}=b-a,{{x}_{2}}=a-b,{{y}_{2}}=a+b\]
So the distance between them
\[\begin{align}
  & =\sqrt{{{\left( a-b-\left( a+b \right) \right)}^{2}}+{{\left( a+b-\left( b-a \right) \right)}^{2}}} \\
 & =\sqrt{{{\left( a-b-a-b \right)}^{2}}+{{\left( a+b-b+a \right)}^{2}}} \\
 & =\sqrt{{{\left( -2b \right)}^{2}}+{{\left( 2a \right)}^{2}}}=\sqrt{4{{b}^{2}}+4{{a}^{2}}}units \\
\end{align}\]
On taking 4 outside the roots, we get as follows:
\[=2\sqrt{{{b}^{2}}+{{a}^{2}}}units\]
Hence the distance between A and B is equal to \[2\sqrt{{{b}^{2}}+{{a}^{2}}}\] units.

(ii) \[A\left( 1,-1 \right),B\left( \dfrac{-1}{2},\dfrac{1}{2} \right)\]
We have \[{{x}_{1}}=1,{{y}_{1}}=-1,{{x}_{2}}=\dfrac{-1}{2},{{y}_{2}}=\dfrac{1}{2}\]
So the distance between A and B
\[\begin{align}
  & =\sqrt{{{\left( \dfrac{-1}{2}-1 \right)}^{2}}+{{\left( \dfrac{1}{2}-\left( -1 \right) \right)}^{2}}} \\
 & =\sqrt{{{\left( \dfrac{-3}{2} \right)}^{2}}+{{\left( \dfrac{1}{2}+1 \right)}^{2}}} \\
 & =\sqrt{\dfrac{9}{4}+\dfrac{9}{4}}=\sqrt{\dfrac{9+9}{4}}=\sqrt{\dfrac{18}{4}}=\dfrac{3}{2}\sqrt{2}units \\
\end{align}\]
Hence the distance between A and B is equal to \[\dfrac{3}{2}\sqrt{2}\] units.

(iii) \[A\left( 8,-2 \right),B\left( 3,-6 \right)\]
We have \[{{x}_{1}}=8,{{y}_{1}}=-2,{{x}_{2}}=3,{{y}_{2}}=-6\]
So the distance between A and B
\[\begin{align}
  & =\sqrt{{{\left( 3-8 \right)}^{2}}+{{\left( -6-\left( -2 \right) \right)}^{2}}} \\
 & =\sqrt{{{\left( -5 \right)}^{2}}+{{\left( -6+2 \right)}^{2}}} \\
 & =\sqrt{25+{{\left( -4 \right)}^{2}}}=\sqrt{25+16}=\sqrt{41}units \\
\end{align}\]
Hence the distance between A and B is equal to \[\sqrt{41}\] units.

(iv) \[A\left( a+b,a-b \right),B\left( a-b,-a-b \right)\]
We have \[{{x}_{1}}=a+b,{{y}_{1}}=a-b,{{x}_{2}}=a-b,{{y}_{2}}=(-a-b)\]
So the distance between A and B
\[\begin{align}
  & =\sqrt{{{\left( a-b-\left( a+b \right) \right)}^{2}}+{{\left( -a-b-\left( a-b \right) \right)}^{2}}} \\
 & =\sqrt{{{\left( a-b-a-b \right)}^{2}}+{{\left( -a-b-a+b \right)}^{2}}} \\
 & =\sqrt{{{\left( -2b \right)}^{2}}+{{\left( -2a \right)}^{2}}}=\sqrt{4{{b}^{2}}+4{{a}^{2}}} \\
 & =2\sqrt{{{a}^{2}}+{{b}^{2}}}units \\
\end{align}\]
Hence the distance between A and B is equal to \[2\sqrt{{{a}^{2}}+{{b}^{2}}}\] units.
Therefore, the distance between all the given points has been calculated.

Note: Sometimes we get confused among \[{{x}_{1}},{{y}_{1}},{{x}_{2}},{{y}_{2}}\] while substituting these values into the formula to find the distance between the points so be careful while substituting the values. Also, remember that distance between any two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is given by \[D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\].