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The points A (-3, b), and B (1, 3) are 5 units apart. Find the value of b.

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Answer:

b = 0

Explanation:

To find the value of b, we will follow the steps below;

Using the distance formula:

D = √(x₂-x₁)² + (y₂-y₁)²

from the question given,

A (-3, b), this implies

(-3, b) = (x₁ ,y₁)

x₁=-3 and y₁ = b

similarly

B (1, 3)

(1, 3) = (x₂,y₂)

this implies

x₂ = 1 and y₂=3

D= 5

we can now proceed to insert the values into the formula and then solve for b

D = √(x₂-x₁)² + (y₂-y₁)²

5 = √(1+3)² + (3-b)²

5 = √4² + (3-b)²

5=√16 + (3-b)²

take the squares of both-side of the equation

5² = 16 + (3-b)²

25 = 16 + (3-b)²

subtract 16 from both-side of the equation

25 - 16 = (3-b)²

9 = (3-b)²

Take the square root of both-side

√9 = 3-b

3 = 3-b

add b to both-side of the equation

3 + b = 3 - b+ b

3 + b = 3

subtract 3 from both-side of the equation

3+b-3 = 3-3

b = 0

Therefore, the value of b is 0

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