if point (X,y) is equidistant from points (a-b,a+b) and (a-b,a+b) prove that bx=ay
Answers
Step-by-step explanation:
Let P(x,y), Q(a+b,b-a) and R(a-b,a+b) be the given points. Then,PQ=PR
Let P(x,y), Q(a+b,b-a) and R(a-b,a+b) be the given points. Then,PQ=PR⇒ {x−(a+b)} 2 +{y−(b−a)} 2
Let P(x,y), Q(a+b,b-a) and R(a-b,a+b) be the given points. Then,PQ=PR⇒ {x−(a+b)} 2 +{y−(b−a)} 2 = {x−(a−b)} 2 +{y−(a+b)} 2
+{y−(a+b)} 2⇒{x−(a+b)} 2 +{y−(b−a)} 2 ={x−(a−b)} 2 +{y−(a+b)} 2
⇒x
⇒x 2 −2x(a+b)+(a+b) 2 +y 2 −2y(b−a)+(b−a)2
⇒x 2 −2x(a+b)+(a+b) 2 +y 2 −2y(b−a)+(b−a)2=x 2 +(a−b) 2 −2x(a−b)+y 2 −2y(a+b)+(a+b) 2
⇒x 2 −2x(a+b)+(a+b) 2 +y 2 −2y(b−a)+(b−a)2=x 2 +(a−b) 2 −2x(a−b)+y 2 −2y(a+b)+(a+b) 2⇒−2x(a+b)−2y(b−a)=−2x(a−b)−2y(a+b)
⇒x 2 −2x(a+b)+(a+b) 2 +y 2 −2y(b−a)+(b−a)2=x 2 +(a−b) 2 −2x(a−b)+y 2 −2y(a+b)+(a+b) 2⇒−2x(a+b)−2y(b−a)=−2x(a−b)−2y(a+b)⇒ax+bx+by−ay=ax−bx+ay+by
⇒x 2 −2x(a+b)+(a+b) 2 +y 2 −2y(b−a)+(b−a)2=x 2 +(a−b) 2 −2x(a−b)+y 2 −2y(a+b)+(a+b) 2⇒−2x(a+b)−2y(b−a)=−2x(a−b)−2y(a+b)⇒ax+bx+by−ay=ax−bx+ay+by⇒2bx=2ay⇒bx=ay
⇒x 2 −2x(a+b)+(a+b) 2 +y 2 −2y(b−a)+(b−a)2=x 2 +(a−b) 2 −2x(a−b)+y 2 −2y(a+b)+(a+b) 2⇒−2x(a+b)−2y(b−a)=−2x(a−b)−2y(a+b)⇒ax+bx+by−ay=ax−bx+ay+by⇒2bx=2ay⇒bx=ayREMARK-We know that a point which is equidistant from point P and Q lies on the
⇒x 2 −2x(a+b)+(a+b) 2 +y 2 −2y(b−a)+(b−a)2=x 2 +(a−b) 2 −2x(a−b)+y 2 −2y(a+b)+(a+b) 2⇒−2x(a+b)−2y(b−a)=−2x(a−b)−2y(a+b)⇒ax+bx+by−ay=ax−bx+ay+by⇒2bx=2ay⇒bx=ayREMARK-We know that a point which is equidistant from point P and Q lies on theperpendicular bisector of PQ. Therefore, bx=ay is the equation of the perpendicular
⇒x 2 −2x(a+b)+(a+b) 2 +y 2 −2y(b−a)+(b−a)2=x 2 +(a−b) 2 −2x(a−b)+y 2 −2y(a+b)+(a+b) 2⇒−2x(a+b)−2y(b−a)=−2x(a−b)−2y(a+b)⇒ax+bx+by−ay=ax−bx+ay+by⇒2bx=2ay⇒bx=ayREMARK-We know that a point which is equidistant from point P and Q lies on theperpendicular bisector of PQ. Therefore, bx=ay is the equation of the perpendicular bisector of PQ
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