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10. Find a relation between x and y such that the point (x, y) is equidistant from the sia
(3,6) and (-3,4).
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10. Find a relation between x and y such that the point (x, y) is equidistant from the sia
(3,6) and (-3,4).
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Answer:
3x + y = 5
Step-by-step explanation:
Using distance formula, which says if there are two points (x₁, y₁) and (x₂, y₂) then the distance between them is √(x₁ - x₂) + (y₁ - y₂).
Here,
⇒ distance b/w (x,y) and (3.6) is equal to the distance b/w (x, y) and (-3,4).
⇒ √(3 - x)² + (6 - y)² = √(-3 - x)² + (4 - y)²
⇒ (3 - x)² + (6 - y)² = (-3-x)² + (4 - y)²
⇒ (3 - x)² - (- 3- x)² = (4 - y)² - (6 - y)²
⇒ (3 - x)² - (3 + x)² = (4 - y)² - (6 - y)²
⇒ (3 - x + 3 + x)(3 - x - 3 - x) = (4 - y + 6 - y)(4 - y - 6 + y)
⇒ 6(- 2x) = (10 - 2y)(- 2)
⇒ - 12x = - 20 + 4y
⇒ - 3x = - 5 + y
⇒ 3x + y = 5
Let,
☃️ According to the question, A is equidistant from B and C .
Where,
Similarly,
Where,
⚡ From Equation 1 :-