Question

please find this according to the class 10​

Answer 1

Answer:

Step-by-step explanation:

If point p is equidistant from points A and B then

AP=BP

applying distance formula -- D=√(x2-x1)² + (y2-y1)²

thus AP= √ (a+b-x)² + (b-a-y)²

and BP= √ (a-b-x)² + (b+a-y)²

thus ,,,

√ (a+b-x)² + (b-a-y)² = √ (a-b-x)² + (b+a-y)²

the root gets cancelled,

(a+b-x)² + (b-a-y)² =  (a-b-x)² + (b+a-y)²

(a+b-x)² - (a-b-x)² = (b+a-y)²- (b-a-y)²

now these terms may be seen as the identity⇒ a²-b²=(a+b)(a-b)

thus,

= (a+b-x+a-b-x)(a+b-x-a+b+x) = (b+a-y+b-a-y)(b+a-y-b+a+y)

= (2a-2x)(2b)= (2b-2y)(2a)

cancelling 2 from both the sides since all is in multiplication

= (a-x)(b)=(b-y)(a)

  ab-bx=ab-ay

ab can get cancelled from both the sides

⇒-bx=-ay

cancelling the negative sign

⇒bx=ay

                      Hence proved!!!!!