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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

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