PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Chapter 5.5, Problem 132E

(a)

To determine

Tograph: The points (3,−1) and (7, 4) .

(a)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The given points are (3,−1) and (7, 4) .

Graph:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 5.5, Problem 132E

Interpretation:

To plot a point on a coordinate plane, the value of x represents the position along the x -axis and the value of y represents the position along the y -axis.

To plot point (3,−1) , move 3 units along the x -axis and 1 unit down.

To plot point (7, 4) , move 7 units along the x -axis and 4 units up.

Therefore, the graph is plotted above.

(b)

To determine

To find:The distance between the points (3,−1) and (7, 4) .

(b)

Expert Solution
Check Mark

Answer to Problem 132E

The distance between the points (3,−1) and (7, 4) is 41 square units.

Explanation of Solution

Given information:

The given points are (3,−1) and (7, 4) .

Calculation:

The distance between two points (x1, y1) and (x2, y2) can be calculated by the distance formula.

  Distance=(x2−x1)2+(y2−y1)2

Let the point (x1, y1) is (3,−1) and (7, 4) is (x2, y2) .

Substitute the values in the above equation to calculate the distance.

  Distance=(7−3)2+(4+1)2=(4)2+52=16+25=41

Therefore, the distance between the points (3,−1) and (7, 4) is 41 .

(c)

To determine

To find:The midpoint of the line segment connecting the points (3,−1) and (7, 4) .

(c)

Expert Solution
Check Mark

Answer to Problem 132E

The midpoint of the line segment is (5, 32) .

Explanation of Solution

Given information:

The given points are (3,−1) and (7, 4) .

Calculation:

The midpoint of the line segment connect the points (x1, y1) and (x2, y2) can be calculated by the midpoint formula.

  Midpoint=(x1+x22, y1+y22)

Let the point (x1, y1) is (3,−1) and (7, 4) is (x2, y2) .

Substitute the values in the above equation to calculate the distance.

  Midpoint=(3+72, 4+(−1)2)=(102, 32)=(5, 32)

Therefore, the midpoint of the line segment is (5, 32) .

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Chapter 5.5, Problem 131E
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Chapter 5.5, Problem 133E

Chapter 5 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

Ch. 5.1 - Prob. 1ECh. 5.1 - Prob. 2ECh. 5.1 - Prob. 3ECh. 5.1 - Prob. 4ECh. 5.1 - Prob. 5ECh. 5.1 - Prob. 6ECh. 5.1 - Prob. 7ECh. 5.1 - Prob. 8ECh. 5.1 - Prob. 9ECh. 5.1 - Prob. 10E
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