If the point (x, y) is equidistant from the points (a + b, b − a) and (a − b, a + b), prove that bx = ay.

Question:

(i) If the point (xy) is equidistant from the points (a + bb − a) and (− b, a + b), prove that bx = ay.

(ii) If the distances of P(x, y) from A(5, 1) and B(–1, 5) are equal then prove that 3x = 2y.

Solution:

(i) As per the question, we have

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

$\Rightarrow(x-a-b)^{2}+(y-b+a)^{2}=(x-a+b)^{2}+(y-a-b)^{2} \quad$ (Squaring both sides)

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

$\Rightarrow-x(a+b)-y(a-b)=-x(a-b)-y(a+b)$

$\Rightarrow-x a-x b-a y+b y=-x a+b x-y a-b y$

$\Rightarrow b y=b x$

Hence, $b x=a y$.

(ii)

As per the question, we have

$\mathrm{AP}=\mathrm{BP}$

$\Rightarrow \sqrt{(x-5)^{2}+(y-1)^{2}}=\sqrt{(x+1)^{2}+(y-5)^{2}}$

$\Rightarrow(x-5)^{2}+(y-1)^{2}=(x+1)^{2}+(y-5)^{2} \quad$ (Squaring both sides)

$\Rightarrow x^{2}-10 x+25+y^{2}-2 y+1=x^{2}+2 x+1+y^{2}-10 y+25$

$\Rightarrow-10 x-2 y=2 x-10 y$

$\Rightarrow 8 y=12 x$

$\Rightarrow 3 x=2 y$

Hence, 3x = 2y.

 

 

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