Question

Find a point within a triangle such that the sum of the square of the distance from the angular point may be minimum

Answer 1

Assume the points are (x1,y1) (x2,y2) (x3,y3)

The point to be determined is (x,y)

F=[(x-x1)^2+(y-y1)^2]+

[(x-x2)^2+(y-y2)^2]+

[(x-x3)^2+(y-y3)^2]

dF/dx=2[x-x1]+2[x-x2]+2[x-x3]=0

 

This will give x=[x1 +x2+x3]/3

 

dF/dx=2[y-y1]+2[y-y2]+2[y-y3]=0

 

This will give y=[y1 +y2+y3]/3

Answer 2

Answer:

The sum of the square of the distance from the angular point may be minimum at  $x=\frac{[x_{1} +x_{2} +x_{3} ]}{3}$ and $y=\frac{y_{1} +y_{2} +y_{3} }{3}$.

Step-by-step explanation:

Given:

a point within a triangle such that the totality of the square of the length from the angular point may exist lowest.

To find:

To discover a point within a triangle such that the totality of the square of the span from the angular point may exist minimum.

Step 1

Consider the points as $(x_{1}, y_{123} ),(x_{2} ,y_{2} ),(x_{3} , y_{3} )$

The point to be defined is $(x,y)$

$F=[(x-x_{1} )^2+(y-y_{1} )^2]+[(x-x_{2} )^2+(y-y_{2} )^2]+[(x-x_{3} )^2+(y-y_{3} )^2]$

$\frac{dF}{dx} =2[x-x_{1} ]+2[x-x_{2} ]+2[x-x_{3} ]=0$

Step 2

This will give,

$x=\frac{[x_{1} +x_{2} +x_{3} ]}{3}$ and

$\frac{dF}{dx} =2[y-y_{1} ]+2[y-y_{2} ]+2[y-y_{3} ]=0$

This will give,

$y=\frac{y_{1} +y_{2} +y_{3} }{3}$

Therefore, the sum of the square of the distance from the angular point may be minimum at

$x=\frac{[x_{1} +x_{2} +x_{3} ]}{3}$ and $y=\frac{y_{1} +y_{2} +y_{3} }{3}$.

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