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Vlad [161]
3 years ago
5

If A( 1,8) B(2,6) and C(4,2) are three points, show that AC=3AB​

Mathematics
2 answers:
Makovka662 [10]3 years ago
8 0

Answer:

<em>Proven below</em>

Step-by-step explanation:

<u>Distance Between Points in the Plane</u>

Given two points A(x,y) and B(w,z), the distance between them is:

d=\sqrt{(w-x)^2+(z-y)^2}

Let's calculate the distance AC, where A(1,8) and C(4,2):

d_{AC}=\sqrt{(4-1)^2+(2-8)^2}

d_{AC}=\sqrt{3^2+(-6)^2}

d_{AC}=\sqrt{9+36}=\sqrt{45}

Since 45=9*5:

d_{AC}=\sqrt{9*5}=3\sqrt{5}

Calculate the distance AB, where A(1,8) and B(2,6)

d_{AB}=\sqrt{(2-1)^2+(6-8)^2}

d_{AB}=\sqrt{1^2+(-2)^2}

d_{AB}=\sqrt{1+4}=\sqrt{5}

It follows that:

d_{AC}=3d_{AB}

faust18 [17]3 years ago
5 0

Answer:

see explanation

Step-by-step explanation:

Calculate AC and AB using the distance formula

d = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = A(1,8) and (x₂, y₂ ) = C(4,2)

AC = \sqrt{(4-1)^2+(2-8)^2}

      = \sqrt{3^2+(-6)^2}

       = \sqrt{9+36}

       = \sqrt{45} = \sqrt{9(5)} = 3\sqrt{5}

Repeat with

(x₁, y₁ ) = A(1, 8) and (x₂, y₂ ) = B(2, 6)

AB = \sqrt{(2-1)^2+(6-8)^2}

      = \sqrt{1^2+(-2)^2}

       = \sqrt{1+4}

       = \sqrt{5}

So AB = \sqrt{5} and AC = 3\sqrt{5}

Thus

AC = 3AB

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M is the MIDPOINT OF FG. FG = 26 AND GM = (3X - 2)
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