Question

(i) If the point (x, y) is equidistant from the points (a + b, b − a) and (a − 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.

Answer 1

(i) As per the question, we have
$$\begin{gathered} \sqrt{{\left(x-a-b\right)}^2+{\left(y-b+a\right)}^2}=\sqrt{{\left(x-a+b\right)}^2+{\left(y-a-b\right)}^2} \\ \Rightarrow {\left(x-a-b\right)}^2+{\left(y-b+a\right)}^2={\left(x-a+b\right)}^2+{\left(y-a-b\right)}^2\left(\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow x^2+{\left(a+b\right)}^2-2x\left(a+b\right)+y^2+{\left(a-b\right)}^2-2y\left(a-b\right)=x^2+{\left(a-b\right)}^2-2x\left(a-b\right)+y^2+{\left(a+b\right)}^2-2y\left(a+b\right) \\ \Rightarrow -x\left(a+b\right)-y\left(a-b\right)=-x\left(a-b\right)-y\left(a+b\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow -xa-xb-ay+by=-xa+bx-ya-by \\ \Rightarrow by=bx \end{gathered}$$
Hence, bx = ay.

(ii)
As per the question, we have
$$\begin{gathered} \mathrm{AP}=\mathrm{BP} \\ \Rightarrow \sqrt{{\left(x-5\right)}^2+{\left(y-1\right)}^2}=\sqrt{{\left(x+1\right)}^2+{\left(y-5\right)}^2} \\ \Rightarrow {\left(x-5\right)}^2+{\left(y-1\right)}^2={\left(x+1\right)}^2+{\left(y-5\right)}^2\left(\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow x^2-10x+25+y^2-2y+1=x^2+2x+1+y^2-10y+25 \end{gathered}$$
$$\begin{gathered} \Rightarrow -10x-2y=2x-10y \\ \Rightarrow 8y=12x \\ \Rightarrow 3x=2y \end{gathered}$$
Hence, 3x = 2y.