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What is the magnitude and direction of with tail and head points M(-3, -2) and N(4, 0)?

Sagot :

DWReadidn

Answer:

Step-by-step explanation:

difference of x-coordinates = 4-(-3) = 7

difference of y-coordinates = 0-(-2)  = 2

length of MN = √(7²+2²) = √53 ≅ 7.28 units

tanθ = 2/7

θ = arctan(2/7) ≅ 15.9°

heading = 74.1° from north

MrRoyalidn

The magnitude and direction of points M(-3, -2) and N(4, 0) is 7.28 units and 74.1 degrees

How to determine the magnitude and direction?

We have:

M(-3, -2) and N(4, 0)

Calculate the magnitude using

[tex]MN =\sqrt{(x_2 -x_1)^2 +(y_2 -y_1)^2[/tex]

This gives

[tex]MN =\sqrt{(4 + 3)^2 +(0 + 2)^2[/tex]

Evaluate

MN = 7.28

The direction is calculated as:

[tex]\theta = 90 - \tan^{-1}(\frac{y_2 - y_1}{x_2 -x_1})[/tex]

This gives

[tex]\theta = 90 -\tan^{-1}(\frac{0+2}{4+3})[/tex]

[tex]\theta =90 - \tan^{-1}(\frac27)[/tex]

Take the arc tan

[tex]\theta = 90 - 15.9\°[/tex]

Evaluate

[tex]\theta = 74.1\°[/tex]

Hence, the magnitude and direction of points M(-3, -2) and N(4, 0) is 7.28 units and 74.1 degrees

Read more about magnitude and direction at:

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