find a relation between X and Y such that the point X Y is equidistant from the point 3,6 and - 3,4
plz answer the question...
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Let P(x , y) , Q( 3, 6) and T( -3 , 4)
A/C to question ,
PQ = PT
√{(x -3)² + (y -6)² } = √{(x +3)² + (y-4)²}
take square both sides,
(x-3)² + ( y -6)² = (x +3)² + (y-4)²
-6x + 9 -12y + 36 = 6x + 9 -8y + 16
-12x -4y + 20 = 0
3x + y = 5
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