Question

Ramu also writes the distance formulas as AB = √((x − x )² + ( y − y )²). Why?​

Answer 1

Step-by-step explanation:

Given,

distance formula as $AB =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Solution:

Let say points $A=(x_{2},y_{1})$

Point B=$=(x_{2},y_{1})$

$AB =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Lets draw a ray AX parallel to X axis

Hence,

any point on AX would be $(x_{1}y_{1})$

Now draw a perpendicular from B at AX meeting at C.

AX is parallel to X-axis hence BC would be parallel to y-axis any point on BC would be $(x_{2},y)$

Intersection point would be $(x_{2},y_{1})$

Length of AC =$x_{2}-x_{1}$

Length of BC =$y_{2}-y_{1}$

Applying Pythagoras theorem in the right angle triangle ACB

$AB^{2}=AC^{2}+CB^{2}$

$→AB^{2}=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

$AB =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Same as distance formula

Hope it helps :-)

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