Question

True or false
If (x,y) is equidistant from (3,6) and (-3,4) then 3x+y= 5

Answer 1

Step-by-step explanation:

Let required point P(x, y) which is equidistant from the point A(3, 6) and B(-3, 4)

So, PA = PB

⇒ PA2 = PB2

⇒ (x – 3)2 + (y – 6)2 = [x -(-3)]2 + (y – 4)2

⇒ x2 + 9 – 6x + y2 + 36 – 12y = (x + 3)2 + (y – 4)2

⇒ x2 + y2 – 6x – 12y + 45 = x2 + 6x + 9 + y2 + 16 – 8y2

⇒ x2 + y2 – 6x – 12y + 45 = x2 + y2 + 6x – 8y + 25

⇒ -6x – 12y = 6x – 8y + 25 – 45

⇒ -6x – 12y – 6x + 8y = -20

⇒ -12x – 4y = -20

⇒ 3x + y = 5

⇒ 3x + y – 5 = 0

Answer 2

Answer:

• True

Explanation:

⌬ Let P (x, y) is equidistant from point A (3,6) and B (-3,4).

Therefore,

➻ PA² = PB²

Formula used:

• Distance Formula = √(x₂ - x₁)² + (y₂ - y₁)²

So,

➻ √(3 - x)² + (6 - y) = √(-3 - x)² + (4 -y)²

Squaring both sides,

➻ (3 - x)² + (6 - y) = (-3 -x)² + (4 - y)²

➻ 9 - 6x + x² + 36 - 12y + y² = 9 + 6x + x² + 16 - 8y + y²

➻ 45 - 6x + x² - 12y + y² = 25 + 6x + x² + y² - 8y

➻ 45 - 25 - 6x - 6x - 12y + 8y = 0

➻ 20 - 12x - 4y = 0

➻ 10 - 6x - 2y = 0

➻ 5 - 3x - y = 0

➻ 3x + y = 5

Hence, Proved!