Question

find a relation between X and y such that the point (x,y) is equidistant from the points (-2,8) and - (3 ,-5)​

Answer 1

Given :

• (-2,8) & (3, - 5) are the two points.

To find :

• Relation between x and y such that the point (x,y) is equidistant from the points (-2,8) and (3 ,-5)

Solution :

Consider the point P(x, y) which is equidistant from the points A(-2, 8) & B(3, -5)

• Solve such problem by applying Distance formula

PA is equal to PB. This is because the point P equidistant from the points A & B

→ PA = PB

• PA = (x,y), (-2, 8)
• PB = (x,y), (3, -5)

• Apply distance formula

→ √(x₂-x₁)² + (y₂-y₁)² = √(x₂-x₁)² + (y₂-y₁)²

→ √(-2 - x)² + (8 - y)² = √(3 - x)² + (-5 - y)²

• Apply identity
• (a - b)² = a² + b² - 2ab

→ √[(-2)² + (x)² - 2 × (-2) × x] + [(8)² + (y)² - 2 × 8 × y] = √[(3)² + (x)² - 2 × 3 × x] + √[(-5)² + (y)² - 2 × (-5) × y]

→ √(4 + x² + 4x) + √(64 + y² - 16y) = √(9 + x² - 6x) + √(25 + y² + 10y)

• Squaring both the sides

→ (4 + x² + 4x) + (64 + y² - 16y) = (9 + x² - 6x) + (25 + y² + 10y)

→ 4 + x² + 4x + 64 + y² - 16y = 9 + x² - 6x + 25 + y² + 10y

→ x² + y² + 4x - 16y + 68 = x² + y² - 6x + 10y + 34

• Cancel x² & y²

→ 4x - 16y + 68 = - 6x + 10y + 34

→ 4x + 6x + 68 - 34 = 10y + 16y

→ 10x + 34 = 26y

→ 10x - 26y + 34 = 0

• Take 2 as a common

→ 2(5x - 13y + 17) = 0

→ 5x - 13y + 17 = 0

→ 5x - 13y = - 17

Hence, 5x - 13y = - 17 is the required relation between x & y

━━━━━━━━━━━━━━━━━━━━━━━━