triangle ABC has vertices at A(1,-3),B(3,0), and C(5,-3). what is the distance from vertex B to the midpoint of AC

Answers

Answer 1

Answer:

The distance from vertex B to the midpoint of AC is 3.

Step-by-step explanation:

From Linear Algebra we understand that location of the midpoint of AC is determined by the following formula:

[tex]M(x,y) = \frac{1}{2}\cdot A(x,y) + \frac{1}{2}\cdot C(x,y)[/tex] (1)

Where:

[tex]A(x,y)[/tex], [tex]C(x,y)[/tex] - Locations of vertices A and C regarding origin, dimensionless.

[tex]M(x,y)[/tex] - Location of the midpoint regarding origin, dimensionless.

If we know that [tex]A(x,y) = (1,-3)[/tex] and [tex]C(x,y) = (5,-3)[/tex], then the midpoint of AC is:

[tex]M(x,y) = \frac{1}{2}\cdot (1,-3)+\frac{1}{2}\cdot (5,-3)[/tex]

[tex]M(x,y) = \left(\frac{1}{2},-\frac{3}{2} \right)+\left(\frac{5}{2}, -\frac{3}{2} \right)[/tex]

[tex]M(x,y) = (3, -3)[/tex]

Lastly, the distance from vertex B to the midpoint of AC is calculated from the Pythagorean Theorem:

[tex]d = \sqrt{(M_{x}-B_{x})^{2}+(M_{y}-B_{y})^{2}}[/tex] (2)

If we know that [tex]M_{x} = 3[/tex], [tex]M_{y} = -3[/tex], [tex]B_{x} = 3[/tex] and [tex]B_{y} = 0[/tex], then the distance is:

[tex]d = \sqrt{(3-3)^{2}+(-3-0)^{2}}[/tex]

[tex]d = 3[/tex]

The distance from vertex B to the midpoint of AC is 3.


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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

c

Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Answers

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Answers

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[tex]\blue{\underline{\underline{\sf{Answer :-}}}} [/tex]

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