Question

A(0, 2), B(1, -0.5), C(2, -3) by distance formula​

Answer 1

Answer:

AB = 5√2.5 units,

BC = √7.25 units,

AC = √29 units.

Step-by-step explanation:

Given points are,

A(0,2), B(1, -0.5), C(2, -3)

Now Consider,

A (0,2) , B (1, -0.5)....., x1 = 0, x2 = 2, y1 = 2, y2 = -0.5

W.K.T,

___________________________________

The distance between 2 points=√(x2-x1)^2 + (y2-y1)^2

___________________________________

= √(2 - 0)^2 + (-0.5 - 2)^2

= √(2)^2 + (-2.5)^2

= √(4) + (6.25)

= √10.25

= 5√2.5

:. AB = 5√2.5 units.

Now Consider,

B(1, -0.5), C(2, -3)..., x1 = 1, x2 = 2, y1 = -0.5, y2 = -3

W.K.T,

___________________________________

The distance between 2 points=√(x2-x1)^2 + (y2-y1)^2

___________________________________

= √[2 - 1]^2 + [-3 -(-0.5)^2]

= √(2-1)^2 + (-3 + 0.5)^2

= √(1)^2 + (-2.5)^2

= √(1) + (6.25)

= √7.25

:. BC = √7.25

Now Consider,

A(0,2), C(2, -3)..., x1 = 0, x2 = 2, y1 = 2, y2 = -3

W.K.T,

___________________________________

The distance between 2 points=√(x2-x1)^2 + (y2-y1)^2

___________________________________

= √(2 - 0)^2 + (-3 -2)^2

= √(2)^2 + (-5)^2

= √(4) + (25)

= √29

:. AC = √29 units.

Hope it Helps!

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