Question

If any point $\text{P}$ is at the equal distances from points $\text{A (a + b, a - b)}$and $\text{B (a - b, a + b),}$ then locus of point $\text{P}$ is

• A. $\text{x – y = 0}$
• B. $\text{ax + by = 0}$
• C. $\text{bx + ay = 0}$
• D. $\text{x + y = 0}$

Answer 1

The correct option is A

$\text{x – y = 0}$

Explanation for the correct option:

Step 1.Take an equidistant point

Let $\text{P(x,y)}$ be the point which is equidistant from points $\text{A (a + b, a – b)}$and $\text{B (a – b, a + b)}$.

Then $\text{PA = PB}$

By using distance formula,

$\sqrt{{[(x-(a+b))}^2+{(y-(a-b))}^2]}$ $\text{=}\sqrt{\left[\right(x-{(a-b))}^2+(y-{(a+b))}^2]}$

Step 2. Squaring on both sides to find the locus of point:

${[(x-(a+b))}^2+{(y-(a-b))}^2]$ $\text{=}\left[\right(x-{(a-b))}^2+(y-{(a+b))}^2]$

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

$\Rightarrow$ $-2x(a+b)–2y(a-b)=–2x(a-b)-2y(a+b)$

$\Rightarrow$ $x(a+b-a+b)+y(a-b-a-b)=0$

$\Rightarrow$ $x\left(2b\right)+y(-2b)=0$

$\Rightarrow$ $2bx–2by=0$

$\Rightarrow$ $x–y=0$

Hence, option $\left(A\right)$ is correct.