Question

Find the distance between the following pair of points
$\left( a{{m}_{1}}^{2},2a{{m}_{1}} \right)$ and $\left( a{{m}_{2}}^{2},2a{{m}_{2}} \right)$

Answer 1

We solve this problem simply by using the distance formula between two points. The distance between the points $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)$ is given as
$\Rightarrow AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
By using the above formula we calculate the distance between the given points by taking out the common terms if any.

Complete step by step answer:
Let us assume that the given points as
$\Rightarrow P=\left( a{{m}_{1}}^{2},2a{{m}_{1}} \right)$
$\Rightarrow Q=\left( a{{m}_{2}}^{2},2a{{m}_{2}} \right)$

We know that the distance formula between two points $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)$ is given as
$\Rightarrow AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
Now, by using the above formula to given two points we get
$\Rightarrow PQ=\sqrt{{{\left( a{{m}_{2}}^{2}-a{{m}_{1}}^{2} \right)}^{2}}+{{\left( 2a{{m}_{2}}-2a{{m}_{1}} \right)}^{2}}}$
By taking the common term from both the terms inside the square root we get
$\Rightarrow PQ=\sqrt{{{a}^{2}}\left[ {{\left( {{m}_{2}}^{2}-{{m}_{1}}^{2} \right)}^{2}}+4{{\left( {{m}_{2}}-{{m}_{1}} \right)}^{2}} \right]}$

We know that the formula of simple algebra that is
$\Rightarrow {{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$
By using this formula in above distance equation we get
$\Rightarrow PQ=\sqrt{{{a}^{2}}\left[ {{\left( {{m}_{2}}-{{m}_{1}} \right)}^{2}}{{\left( {{m}_{1}}+{{m}_{2}} \right)}^{2}}+4{{\left( {{m}_{2}}-{{m}_{1}} \right)}^{2}} \right]}$
Now by taking the common terms out and separating the terms we get
$\Rightarrow PQ=\sqrt{{{\left[ a\left( {{m}_{2}}-{{m}_{1}} \right) \right]}^{2}}}\sqrt{{{\left( {{m}_{2}}+{{m}_{1}} \right)}^{2}}+4}.........equation(i)$
We know the standard result that is
$\Rightarrow \sqrt{{{x}^{2}}}=\pm x$
But we know that the distance cannot be negative
So, we can take the above standard result as
$\Rightarrow \sqrt{{{x}^{2}}}=\left| x \right|$
By using the above result in the equation (i) we get
$\Rightarrow PQ=\left| a\left( {{m}_{2}}-{{m}_{1}} \right) \right|\sqrt{{{\left( {{m}_{2}}+{{m}_{1}} \right)}^{2}}+4}$
We know that the formula of square of sum of two numbers as
$\Rightarrow {{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$
By using this formula in above equation we get
$\Rightarrow PQ=\left| a\left( {{m}_{2}}-{{m}_{1}} \right) \right|\sqrt{{{m}_{2}}^{2}+2{{m}_{2}}{{m}_{1}}+{{m}_{1}}^{2}+4}$

Therefore the distance between the given pair of points is $\left| a\left( {{m}_{2}}-{{m}_{1}} \right) \right|\sqrt{{{m}_{2}}^{2}+2{{m}_{2}}{{m}_{1}}+{{m}_{1}}^{2}+4}$

Note: Students may make mistakes in taking the terms out of square root.
We know the standard result that is
$\Rightarrow \sqrt{{{x}^{2}}}=\pm x$
But we know that the distance cannot be negative
So, we can take the above standard result as
$\Rightarrow \sqrt{{{x}^{2}}}=\left| x \right|$
But students may miss this point and take the value as
$\Rightarrow \sqrt{{{x}^{2}}}=\pm x$
Or else they can take simply
$\Rightarrow \sqrt{{{x}^{2}}}=x$
This gives the wrong answer.
Because we don’t know about the sign of $'x'$
So, the above mentioned point has to be taken care of.