# Find the distances between the following pair of points $(a,0) $and$(0,b)$.

Last updated date: 20th Mar 2023

•

Total views: 306.6k

•

Views today: 6.86k

Answer

Verified

306.6k+ views

Hint: Use distance formula. Use$({{x}_{1}},{{y}_{1}})\equiv (a,0)$ and $({{x}_{2}},{{y}_{2}})\equiv (0,b)$. So substitute the points you will get the answer.

So about distance,

Distance is the total movement of an object without any regard to direction. We can define

distance as to how much ground an object has covered despite its starting or ending point.

So let $A$ and $B$ be the points on a graph above.

So we have to find the distance between these points $AB$.

We can run lines down from $A$, and along from $B$, to make a Right Angled Triangle.

And with a little help from Pythagoras we know that,

${{a}^{2}}+{{b}^{2}}={{c}^{2}}$

So we get,

By simplifying,

$c=\sqrt{{{a}^{2}}+{{b}^{2}}}$…………. (1)

Now let the coordinates of $A({{x}_{1}},{{y}_{1}})$ and $B({{x}_{2}},{{y}_{2}})$

${{x}_{1}}$ means the $x$ coordinate of point $A$ and ${{y}_{1}}$ means the $y$ coordinate

of point $A$,

${{x}_{2}}$ means the $x$ coordinate of point $B$ and ${{y}_{2}}$ means the $y$ coordinate

of point $B$,

So the horizontal distance $a$ is $({{x}_{1}}-{{x}_{2}})$,

Also the vertical distance $b$ is $({{y}_{1}}-{{y}_{2}})$.

Now we can solve for $c$ (the distance between the points),

So from (1), Substituting the value of $a$ and $b$,

We get,

$c=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$

So we can write$c$ as $dist.(AB)$,

So we get,

$dist.(AB)=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$

This formula finds the length of a line that stretches between two points: Point $A$ and Point $B$. The linear distance is the square root of the square of the horizontal distance plus

the square of the vertical distance between two points

So the above we get the distance.

So the above formula is a distance formula.

So now we have to find the distance between two points i.e.$(a,0)$ and $(0,b)$,

So let$ A(a,0)$ and $B(0,b)$,

So here ${{x}_{1}}=a,{{x}_{2}}=0,{{y}_{1}}=0 $and \[{{y}_{2}}=b\],

So Using distance formula, we get,

$dist.(AB)=\sqrt{{{(a-0)}^{2}}+{{(0-b)}^{2}}}$

So simplifying in simple manner we get,

$dist.(AB)=\sqrt{{{a}^{2}}+{{b}^{2}}}$

So we get the final distance between the points$(a,0)$and$(0,b)$as$\sqrt{{{a}^{2}}+{{b}^{2}}}$.

Note: So be familiar with the distance formula i.e. $dist.(AB)=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$. So if the question is

asked to find the distance so first equate the points with$A({{x}_{1}},{{y}_{1}})$ and $B({{x}_{2}},{{y}_{2}})$. Then it will get easy to solve the problem. Otherwise confusion also

occurs between the points. The points get interchange so equate the points to avoid the confusion.

So about distance,

Distance is the total movement of an object without any regard to direction. We can define

distance as to how much ground an object has covered despite its starting or ending point.

So let $A$ and $B$ be the points on a graph above.

So we have to find the distance between these points $AB$.

We can run lines down from $A$, and along from $B$, to make a Right Angled Triangle.

And with a little help from Pythagoras we know that,

${{a}^{2}}+{{b}^{2}}={{c}^{2}}$

So we get,

By simplifying,

$c=\sqrt{{{a}^{2}}+{{b}^{2}}}$…………. (1)

Now let the coordinates of $A({{x}_{1}},{{y}_{1}})$ and $B({{x}_{2}},{{y}_{2}})$

${{x}_{1}}$ means the $x$ coordinate of point $A$ and ${{y}_{1}}$ means the $y$ coordinate

of point $A$,

${{x}_{2}}$ means the $x$ coordinate of point $B$ and ${{y}_{2}}$ means the $y$ coordinate

of point $B$,

So the horizontal distance $a$ is $({{x}_{1}}-{{x}_{2}})$,

Also the vertical distance $b$ is $({{y}_{1}}-{{y}_{2}})$.

Now we can solve for $c$ (the distance between the points),

So from (1), Substituting the value of $a$ and $b$,

We get,

$c=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$

So we can write$c$ as $dist.(AB)$,

So we get,

$dist.(AB)=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$

This formula finds the length of a line that stretches between two points: Point $A$ and Point $B$. The linear distance is the square root of the square of the horizontal distance plus

the square of the vertical distance between two points

So the above we get the distance.

So the above formula is a distance formula.

So now we have to find the distance between two points i.e.$(a,0)$ and $(0,b)$,

So let$ A(a,0)$ and $B(0,b)$,

So here ${{x}_{1}}=a,{{x}_{2}}=0,{{y}_{1}}=0 $and \[{{y}_{2}}=b\],

So Using distance formula, we get,

$dist.(AB)=\sqrt{{{(a-0)}^{2}}+{{(0-b)}^{2}}}$

So simplifying in simple manner we get,

$dist.(AB)=\sqrt{{{a}^{2}}+{{b}^{2}}}$

So we get the final distance between the points$(a,0)$and$(0,b)$as$\sqrt{{{a}^{2}}+{{b}^{2}}}$.

Note: So be familiar with the distance formula i.e. $dist.(AB)=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$. So if the question is

asked to find the distance so first equate the points with$A({{x}_{1}},{{y}_{1}})$ and $B({{x}_{2}},{{y}_{2}})$. Then it will get easy to solve the problem. Otherwise confusion also

occurs between the points. The points get interchange so equate the points to avoid the confusion.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE