Question

If the point $\left( x,y \right)$ be equidistant from the point $\left( a+b,b-a \right)$ and $\left( a-b,a+b \right)$ prove that $bx=ay$ .

Answer 1

Hint: e will assume points C be $\left( x,y \right)$ , A be $\left( a+b,b-a \right)$ and B be $\left( a-b,a+b \right)$ . So, we can say that $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ . Then we will use the distance formula $\sqrt{{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}}$ for $CA=CB$ . So, using this formula and on substituting the values and solving we will be proving $bx=ay$ . We will use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ .

Complete step-by-step answer:
Here, we will draw the figure for this where one point is equidistant from the other two points. So, we get as

Let point C be $\left( x,y \right)$ , A be $\left( a+b,b-a \right)$ and B be $\left( a-b,a+b \right)$ . So, we will take $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ .
Now, it is given that $CA=CB$ . So, we will use distance formula which is given as $\sqrt{{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}}$ . So, using this we get as
$CA=CB$
Applying formula for CA and CB points i.e. $\left( x,y \right)$ , $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( x,y \right)$ , $\left( {{x}_{2}},{{y}_{2}} \right)$ . So, we get as
$\Rightarrow \sqrt{{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}}=\sqrt{{{\left( x-{{x}_{2}} \right)}^{2}}+{{\left( y-{{y}_{2}} \right)}^{2}}}$
On putting values, we get as
$\Rightarrow \sqrt{{{\left( x-\left( a+b \right) \right)}^{2}}+{{\left( y-\left( b-a \right) \right)}^{2}}}=\sqrt{{{\left( x-\left( a-b \right) \right)}^{2}}+{{\left( y-\left( a+b \right) \right)}^{2}}}$
Now, we will square on both sides and remove the square root. So, we get as
$\Rightarrow {{\left( x-\left( a+b \right) \right)}^{2}}+{{\left( y-\left( b-a \right) \right)}^{2}}={{\left( x-\left( a-b \right) \right)}^{2}}+{{\left( y-\left( a+b \right) \right)}^{2}}$
Now, we apply the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ . So, we can write the equation as
$\begin{align}
  & \Rightarrow {{x}^{2}}+{{\left( a+b \right)}^{2}}-2x\left( a+b \right)+{{y}^{2}}+{{\left( b-a \right)}^{2}}-2y\left( b-a \right)= \\
 & \text{ }{{x}^{2}}+{{\left( a-b \right)}^{2}}-2x\left( a-b \right)+{{y}^{2}}+{{\left( a+b \right)}^{2}}-2y\left( a+b \right) \\
\end{align}$
On further cancelling the ${{x}^{2}}$ , ${{y}^{2}}$ from both sides, we get as
$\begin{align}
  & \Rightarrow {{\left( a+b \right)}^{2}}-2x\left( a+b \right)+{{\left( b-a \right)}^{2}}-2y\left( b-a \right)= \\
 & \text{ }{{\left( a-b \right)}^{2}}-2x\left( a-b \right)+{{\left( a+b \right)}^{2}}-2y\left( a+b \right) \\
\end{align}$
Now, if we take square terms on one side and remaining terms on other side, we get as
$\begin{align}
  & \Rightarrow {{\left( a+b \right)}^{2}}+{{\left( b-a \right)}^{2}}-{{\left( a-b \right)}^{2}}-{{\left( a+b \right)}^{2}}= \\
 & 2x\left( a+b \right)+2y\left( b-a \right)-2x\left( a-b \right)-2y\left( a+b \right) \\
\end{align}$
On cancelling plus minus terms from both sides, we get as
$\Rightarrow {{\left( b-a \right)}^{2}}-{{\left( a-b \right)}^{2}}=2x\left( a+b \right)+2y\left( b-a \right)-2x\left( a-b \right)-2y\left( a+b \right)$
On expanding Left hand side terms, we get as
$\begin{align}
  & \Rightarrow {{a}^{2}}+{{b}^{2}}-2ab-\left( {{a}^{2}}+{{b}^{2}}-2ab \right)= \\
 & 2x\left( a+b \right)+2y\left( b-a \right)-2x\left( a-b \right)-2y\left( a+b \right) \\
\end{align}$
Thus, on simplification and rearranging terms we get equation as
$\Rightarrow 2x\left( a+b \right)+2y\left( b-a \right)=2x\left( a-b \right)+2y\left( a+b \right)$
On cancelling 2 from both sides and multiplying the brackets we get as
$\Rightarrow ax+bx+by-ay=ax-bx+ay+by$
On cancelling the same terms on both sides, we get as
$\Rightarrow bx-ay=-bx+ay$
$\Rightarrow 2bx=2ay$
Thus, we get as
$\Rightarrow bx=ay$
Hence, proved.

Note: Students should understand the question and then draw the figure accordingly then only we can know that $CA=CB$ . Also, equidistant formulas should be known, then only by solving the equation we will get the answer. Be careful while solving long equations if there is a single sign mistake then the answer will not come, and students will get confused. So, please avoid this mistake.