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High School Math Algebra High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

Prove that the triangle with given vertices is a right triangle. Given: Vertices are 3 , 5 , − 2 , 6 , 1 , 3 . Formula Used: Slope of a line is given by m = y 2 − y 1 x 2 − x 1 . Slopes of perpendicular lines are negative reciprocal of each other. Calculation: The given vertices of the triangle are 3 , 5 , − 2 , 6 , 1 , 3 . The slope of the line joining the points 3 , 5 , − 2 , 6 is m 1 = 6 − 5 − 2 − 3 = − 1 5 Now, the slope of the line with points − 2 , 6 , 1 , 3 is m 2 = 3 − 6 1 + 2 = − 1 Now, the slope of the line with points 3 , 5 , 1 , 3 is m 3 = 3 − 5 1 − 3 = 1 Since, the slopes m 2 and m 2 are negative reciprocal of each other. Hence, the lines joining these points are perpendicular to each other. Therefore, we can conclude that the given vertices are of a right triangle.

Prove that the triangle with given vertices is a right triangle. Given: Vertices are 3 , 5 , − 2 , 6 , 1 , 3 . Formula Used: Slope of a line is given by m = y 2 − y 1 x 2 − x 1 . Slopes of perpendicular lines are negative reciprocal of each other. Calculation: The given vertices of the triangle are 3 , 5 , − 2 , 6 , 1 , 3 . The slope of the line joining the points 3 , 5 , − 2 , 6 is m 1 = 6 − 5 − 2 − 3 = − 1 5 Now, the slope of the line with points − 2 , 6 , 1 , 3 is m 2 = 3 − 6 1 + 2 = − 1 Now, the slope of the line with points 3 , 5 , 1 , 3 is m 3 = 3 − 5 1 − 3 = 1 Since, the slopes m 2 and m 2 are negative reciprocal of each other. Hence, the lines joining these points are perpendicular to each other. Therefore, we can conclude that the given vertices are of a right triangle.

Solution Summary: The author demonstrates that the triangle with given vertices is a right triangle. The slope of perpendicular lines is negative reciprocal of each other.

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

15th Edition

ISBN: 9780133281163

Author: Prentice Hall

Publisher: Prentice Hall

See similar textbooks

Expert Solution & Answer

Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).

Chapter 2.4, Problem 55PPSE

Solution

Prove that the triangle with given vertices is a right triangle.

Given:

Vertices are 3,5,−2,6,1,3 .

Formula Used:

Slope of a line is given by m=y2−y1x2−x1 .

Slopes of perpendicular lines are negative reciprocal of each other.

Calculation:

The given vertices of the triangle are 3,5,−2,6,1,3 .

The slope of the line joining the points 3,5,−2,6 is

m1=6−5−2−3=−15

Now, the slope of the line with points −2,6,1,3 is

m2=3−61+2=−1

Now, the slope of the line with points 3,5,1,3 is

m3=3−51−3=1

Since, the slopes m2 and m2 are negative reciprocal of each other.

Hence, the lines joining these points are perpendicular to each other.

Therefore, we can conclude that the given vertices are of a right triangle.

Chapter 2.4, Problem 54PPSE

Chapter 2.4, Problem 56PPSE

Chapter 2 Solutions

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

Ch. 2.1 - Prob. 1LC Ch. 2.1 - Prob. 2LC Ch. 2.1 - Prob. 3LC Ch. 2.1 - Prob. 4LC Ch. 2.1 - Prob. 5LC Ch. 2.1 - Prob. 6LC Ch. 2.1 - Prob. 7LC Ch. 2.1 - Prob. 8PPSE Ch. 2.1 - Prob. 9PPSE Ch. 2.1 - Prob. 10PPSE

Ch. 2.1 - Prob. 11PPSE Ch. 2.1 - Prob. 12PPSE Ch. 2.1 - Prob. 13PPSE Ch. 2.1 - Prob. 14PPSE Ch. 2.1 - Prob. 15PPSE Ch. 2.1 - Prob. 16PPSE Ch. 2.1 - Prob. 17PPSE Ch. 2.1 - Prob. 18PPSE Ch. 2.1 - Prob. 19PPSE Ch. 2.1 - Prob. 20PPSE Ch. 2.1 - Prob. 21PPSE Ch. 2.1 - Prob. 22PPSE Ch. 2.1 - Prob. 23PPSE Ch. 2.1 - Prob. 24PPSE Ch. 2.1 - Prob. 25PPSE Ch. 2.1 - Prob. 26PPSE Ch. 2.1 - Prob. 27PPSE Ch. 2.1 - Prob. 28PPSE Ch. 2.1 - Prob. 29PPSE Ch. 2.1 - Prob. 30PPSE Ch. 2.1 - Prob. 31PPSE Ch. 2.1 - Prob. 32PPSE Ch. 2.1 - Prob. 33PPSE Ch. 2.1 - Prob. 34PPSE Ch. 2.1 - Prob. 35PPSE Ch. 2.1 - Prob. 36PPSE Ch. 2.1 - Prob. 37PPSE Ch. 2.1 - Prob. 38PPSE Ch. 2.1 - Prob. 39PPSE Ch. 2.1 - Prob. 40PPSE Ch. 2.1 - Prob. 41PPSE Ch. 2.1 - Prob. 42PPSE Ch. 2.1 - Prob. 43PPSE Ch. 2.1 - Prob. 44PPSE Ch. 2.1 - Prob. 45PPSE Ch. 2.1 - Prob. 46PPSE Ch. 2.1 - Prob. 47PPSE Ch. 2.1 - Prob. 48PPSE Ch. 2.1 - Prob. 49PPSE Ch. 2.1 - Prob. 50PPSE Ch. 2.1 - Prob. 51PPSE Ch. 2.1 - Prob. 52PPSE Ch. 2.2 - Prob. 1LC Ch. 2.2 - Prob. 2LC Ch. 2.2 - 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