Question

If the point (2,1) and (1,-2) are equidistant from the point x,y showt that x+3y=0

Answer 1

let the point is (x,y) from which the point 2,1 and 1,-2 are eqidistant then by distance formula we equate both ..i.e
$(x - 2) { }^{2} + (y - 1) {}^{2} = (x - 1) {?}^{2} + (y + 2) ^{2}$
from here we get our answer

Answer 2

hello users ....

solution:-
we know that;
According to distance formula
Distance between points A(a,b) And B(c,d)
= √ [ (c - a)² + (d - b)² ] 

Here,
distance between point A(2,1) And X(x,y) = distance between points B(1,-2) And X(x,y) 

=> √ [(x-2)² + (y-1)²]  =  √ [(x-1)² + (y+2)²]

=> (x-2)² + (y-1)²  =  (x-1)² + (y+2)²

=> ( x² + 4 - 4x ) + ( y² + 1 - 2y ) = ( x² + 1 - 2x) + ( y² + 4 + 4y)

=> x² + 4 - 4x +  y² + 1 - 2y =  x² + 1 - 2x + y² + 4 + 4y

=> x² + 4 - 4x +  y² + 1 - 2y - x² - 1 + 2x - y² - 4 - 4y = 0 

=> (x² - x² ) + ( y² - y²) + (4 - 4+1-1) + ( -4x + 2x) + ( -2y - 4y) = 0

=> -2x - 6y = 0 

=> x + 3y = 0 

Hence ; 
Showed...

# hope it helps  :)