Question

The locus of the points equidistant from the points (a+b,a-b) and (a-b,a+b)
1) bx - ay = 0
2) bx + ay = 0
3) ax-by = 0
4) x-y=0​

Answer 1

Answer- The above question is from the chapter 'Straight Lines'.

Let's know about locus first.

Locus: It is the path traced by a point moving under some set of geometrical conditions.

Every path has some equation.

To solve the questions related to locus,

1. Put the point whose locus we want to find as (x, y) or (h, k).

2. Apply all geometrical conditions given.

3. Eliminate all variables in locus problem.

Given question: The locus of the points equidistant from the points (a + b, a - b) and (a - b, a + b) is _______ .

1) bx - ay = 0

2) bx + ay = 0

3) ax  -by = 0

4) x - y = 0​

Solution:

STEP - 1: Put the point whose locus we want to find as (x, y) or (h, k).

Let P(x, y) be the point equidistant from the points A (a + b, a - b) and B (a - b, a + b)

STEP - 2: Apply all geometrical conditions given.

PA = PB

Squaring both sides, we get,

PA² = PB²

Using distance formula, we get,

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

Solving it, we get,

x² + (a + b)² - 2x(a + b) + y² + (a - b)² - 2y(a - b) = x² + (a - b)² - 2x(a - b) + y² + (a + b)² - 2y(a + b)

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

ax + bx + ay - by = ax - bx + ay + by

bx + bx - by - by = 0

2bx - 2by = 0

2b(x - y) = 0

⇒ x - y = 0

∴ 4)  x - y = 0​ is the correct option.