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2 years ago

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1. Find the slope of the line that goes through the points (2, 5) and (6, 7)

1 answer:

ella [17]2 years ago

7 0

Answer:

$\displaystyle m = \frac{1}{2}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Algebra I</u>

Coordinates (x, y)
Slope Formula: $\displaystyle m = \frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Point (2, 5)

Point (6, 7)

<u>Step 2: Find slope </u><em><u>m</u></em>

Substitute in points [Slope Formula]: $\displaystyle m = \frac{7-5}{6-2}$
[Fraction] Subtract: $\displaystyle m = \frac{2}{4}$
[Fraction] Simplify: $\displaystyle m = \frac{1}{2}$

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Step-by-step explanation:

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$B = (2,8) \to (x_2,y_2)$

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Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

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$B = (2,8) \to (x_2,y_2)$

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$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

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$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

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

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

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$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

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$\cos A =\frac{-18}{-68.42}$

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$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

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Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

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