Question

a car is located at the point (-6,-4) in xy plane. A bottle of milk is kept at (5,11). The cat wosh to consume the milk travelling through shortest possible distance. Find the equation of thr pads it needs to take its milk

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Answer 1

Given :

• A car is located at the point (-6,-4) in xy plane.
• A bottle of milk is kept at (5,11).

To Find :

• Equation of the path it needs to take its milk

Solution :

So let's take A as,

Points = (-6,-4)

• x₁ = -6 , y₁ = -4
• A = ( x₁ , y₁ ) = ( -6 , -4 )

So let's take B as,

Points = (5,11)

• x₂ = 5 , y₂ = 11
• B = ( x₂ , y₂ ) = ( 5 , 11 )

Shortest path between A and B is a line joining A and B.

So we've y₂ = 11, y₁ = -4, x₂ = 5, x₁ = -6

By using formula,

⇒ m = y₂ - y₁ / x₂ - x₁

⇒ m = 11 - ( -4 ) / 5 - ( -6 )

⇒ m = 11 + 4 / 5 + 6

⇒ m = 15 / 11

y - y₁ = m ( x - x₁ )

⇒ y - ( -4 ) = 15 / 11 ( x - ( -6) )

⇒ ( y + 4 ) × 11 = 15 × ( x - ( -6 ) )

⇒ 11y + 44 = 15x + 90

⇒ 15x - 11y + 46 = 0

•°• 15x - 11y + 46 = 0 is the equation of the path it needs to take its milk.

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Answer 2

Answer:

$\huge\mathfrak\red{Given :}$

A car is located at the point (-6,-4) in xy plane.

A bottle of milk is kept at (5,11).

$\huge\mathfrak\blue{To Find :}$

Equation of the path it needs to take its milk

$\huge\mathfrak\pink{Solution :}$

So let's take A as,

Points = (-6,-4)

x₁ = -6 , y₁ = -4

A = ( x₁ , y₁ ) = ( -6 , -4 )

So let's take B as,

Points = (5,11)

x₂ = 5 , y₂ = 11

B = ( x₂ , y₂ ) = ( 5 , 11 )

Shortest path between A and B is a line joining A and B.

So we've y₂ = 11, y₁ = -4, x₂ = 5, x₁ = -6

$\huge\mathfrak\green{By ..using.. formula,}$

⇒ m = y₂ - y₁ / x₂ - x₁

⇒ m = 11 - ( -4 ) / 5 - ( -6 )

⇒ m = 11 + 4 / 5 + 6

⇒ m = 15 / 11

y - y₁ = m ( x - x₁ )

⇒ y - ( -4 ) = 15 / 11 ( x - ( -6) )

⇒ ( y + 4 ) × 11 = 15 × ( x - ( -6 ) )

⇒ 11y + 44 = 15x + 90

⇒ 15x - 11y + 46 = 0

•°• 15x - 11y + 46 = 0 is the equation of the path it needs to take its milk.

_____________________________