Question

If the point P( x,y ) is equidistant from the point Q( a+b , b-a ) and R( a-b , a+b ) then prove that bx=ay
Plz solve if u know how to solve the problem because my boards are there.

Answer 1

Answer:

Step-by-step explanation:

PA=PB

take square both side

PA^2=PB^2

now use distance

formula ,

{x-(a+b)}^2+{y-(b-a)}^2={x-(a-b)}^2+{y-(a+b)}^2

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

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

=>2x{a-b-a-b}=2y{b-a-a-b}

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

=>bx=ay

hence proved

kindly mark me the brainliest

Answer 2

given

P(x,y) is equidistant from the point Q(a+b,a-b) and R(a-b,a+b)

To find, prove that bx = ay.

∴ PA^{2} =PB^{2}

Using distance formula,

\sqrt{(x_{2}- x_{1})^{2}+(y_{2}- y_{1})^{2}}

⇒{(a+b-x)^{2}+{(b-a-y)^{2}}={(a-b-x)^{2}+{(a+b-y)^{2}}

⇒{(a+b-x)^{2}-(a-b-x)^{2}={(a+b-y)^{2}-(b-a-y)^{2}

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

⇒ (2a-2x)(2b)=(2b-2y)(2a)

⇒ (a-x)(b)=(b-y)(a)

⇒ ab-bx=ab-ay

⇒bx=ay, proved.