Question

The x-coordinate of point P is twice is y coordinate. If P is equidistant from Q ( 2, - 5 ) and R ( - 3, 6 ), find the coordinates of P.

Answer 1

Answer:

Coordinates of p is (16,8)

Step-by-step explanation:

Answer 2

Hey there !! 


The given points be Q( 2 , -5 ) and R( -3 , 6 ) .

And, let the required point be P( x , y ) .

Then,

°•° x = 2y .

▶Now,

→ PQ = PR . [ A and B are equidistant from P ] 

[ Squaring both side ] 

=> PQ² = PR² .

[ Using distance formula ] 

$\begin{lgathered}= > {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} = {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} . \\ \\\end{lgathered}$

=> ( x - 2 )² + ( y + 5 )² = ( x + 3 )² + ( y - 6 )² .

=> x² + 4 - 4x + y² + 25 + 10y = x² + 9 + 6x + y² + 36 - 12y .

=> 29 - 4x + 10y = 45 + 6x - 12y .

=> 10y + 12y = 45 - 29 + 6x + 4x .

=> 22y = 16 + 10x .

=> 22y = 16 + 10(2y) .... [ °•° x = 2y ] .

=> 22y = 16 + 20y .

=> 22y - 20y = 16 .

=> 2y = 16 .

=> y = 16/2 .

•°• y = 8 .


•°• x = 2y = 2 × 8 = 16 .



✔✔ Hence, coordinates of point P is ( 16 , 8 ) .✅✅




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