a. Answer 3 x + 4 y = 7 Explanation Given information: Coordinates A ( − 3 , 4 ) , B ( 5 , − 2 ) Formula used: y − y 1 = ( y 2 − y 1 ) ( x 2 − x 1 ) ( x − x 1 ) Calculation: Let the coordinates point of A ( − 3 , 4 ) is x 1 , y 1 and B ( 5 , − 2 ) is x 2 , y 2 respectively Substituting the values of x 1 , y 1 , x 2 , y 2 in the formula y − 4 = ( − 2 − 4 ) ( 5 + 3 ) ( x + 3 ) Calculating for the equation y − 4 = ( − 3 ) ( 4 ) ( x + 3 ) Using cross multiplication method 4 y − 16 = − 3 x − 9 Taking all the variables in the left hand side and constant term in the right hand side 3 x + 4 y = 7 b. To determine To find the equation of a line that is parallel to the line shown and passes through the point ( − 7 , − 6 ) Answer 3 x + 4 y = − 45 Explanation Given information: Equation of the line shown is 3 x + 4 y = 7 Points ( − 7 , − 6 ) = x 1 , y 1 Formula used: Use post slope formula to find the equation y − y 1 = m ( x − x 1 ) To calculate m y = m x + c Calculation: Substituting the values to calculate m 4 y = 7 − 3 x Comparing this equation to the formula to calculate m y = 7 4 − 3 4 x The value of m is m = − 3 4 Now finding equation using post slope formula and substituting values ( y + 6 ) = − 3 4 ( x + 7 ) Using cross multiplication method 4 y + 24 = − 3 x − 21 The equation of the line passing through the points is 3 x + 4 y = − 45 c. To determine To find the equation of a line that is perpendicular to the line shown and passes through the point ( − 3 , 4 ) Answer 4 x − 3 y = − 24 Explanation Given information: Equation of the line shown is 3 x + 4 y = 7 Value of m 1 = − 3 4 Points ( − 3 , 4 ) Formula used: Use post slope formula to find the equation y − y 1 = m ( x − x 1 ) To calculate m m 1 × m 2 = − 1 Calculation: Calculating for m − 3 4 × m 2 = − 1 Final value is m 2 = − 3 4 Perpendicular slope must be opposite reciprocal of each other m 2 = 4 3 Substituting values in post slope formula ( y − 4 ) = 4 3 ( x + 3 ) Calculating the equation 4 x − 3 y = − 24 d. To determine To find the equation of a line that is perpendicular to the line shown and passes through the point ( 5 , − 2 ) Answer 4 x − 3 y = 26 Explanation Given information: Equation of the line shown is 3 x + 4 y = 7 Points ( 5 , − 2 ) Formula used: Use post slope formula to find the equation y − y 1 = m ( x − x 1 ) Calculation: substituting the values in the formula ( y + 2 ) = 4 3 ( x − 5 ) Calculating the equation 4 x − 3 y = 26 e. To determine To Graph : To graph the given lines Explanation Given information : Given equations 3 x + 4 y = 7 , 3 x + 4 y = − 45 , 4 x − 3 y = − 24 , 4 x − 3 y = 26 Graph : Interpretation : This graph shows that it is forming a Quadrilateral

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Videos

Question

Chapter 10, Problem 15PTTS

To determine

To find the equation of the line AB

Expert Solution

Answer to Problem 15PTTS

3x+4y=7

Explanation of Solution

Given information:

Coordinates A(−3,4) , B(5,−2)

Formula used:

y−y1=(y2−y1)(x2−x1)(x−x1)

Calculation:

Let the coordinates point of A(−3,4) is x1 , y1 and B(5,−2) is x2 , y2 respectively

Substituting the values of x1 , y1 , x2 , y2 in the formula

y−4=(−2−4)(5+3)(x+3)

Calculating for the equation

y−4=(−3)(4)(x+3)

Using cross multiplication method

4y−16=−3x−9

Taking all the variables in the left hand side and constant term in the right hand side

3x+4y=7

To determine

To find the equation of a line that is parallel to the line shown and passes through the point (−7,−6)

Expert Solution

Answer to Problem 15PTTS

3x+4y=−45

Explanation of Solution

Given information:

Equation of the line shown is 3x+4y=7

Points

(−7,−6)=x1,y1

Formula used:

Use post slope formula to find the equation

y−y1=m(x−x1)

To calculate m y=mx +c

Calculation:

Substituting the values to calculate m

4y=7−3x

Comparing this equation to the formula to calculate m

y=74−34x

The value of m is

m=−34

Now finding equation using post slope formula and substituting values

(y+6)=−34(x+7)

Using cross multiplication method

4y+24=−3x−21

The equation of the line passing through the points is

3x+4y=−45

To determine

To find the equation of a line that is perpendicular to the line shown and passes through the point (−3,4)

Expert Solution

Answer to Problem 15PTTS

4x−3y=−24

Explanation of Solution

Given information:

Equation of the line shown is 3x+4y=7

Value of

m1=−34

Points

(−3,4)

Formula used:

Use post slope formula to find the equation

y−y1=m(x−x1)

To calculate m

m1×m2=−1

Calculation:

Calculating for m

−34×m2=−1

Final value is

m2=−34

Perpendicular slope must be opposite reciprocal of each other

m2=43

Substituting values in post slope formula

(y−4)=43(x+3)

Calculating the equation

4x−3y=−24

To determine

To find the equation of a line that is perpendicular to the line shown and passes through the point (5,−2)

Expert Solution

Answer to Problem 15PTTS

4x−3y=26

Explanation of Solution

Given information:

Equation of the line shown is 3x+4y=7

Points

(5,−2)

Formula used:

Use post slope formula to find the equation

y−y1=m(x−x1)

Calculation: substituting the values in the formula

(y+2)=43(x−5)

Calculating the equation

4x−3y=26

To determine

To Graph :

To graph the given lines

Expert Solution

Explanation of Solution

Given information :

Given equations

3x+4y=7 , 3x+4y=−45 , 4x−3y=−24 , 4x−3y=26

Graph :

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 10, Problem 15PTTS

Interpretation :

This graph shows that it is forming a Quadrilateral

Chapter 10, Problem 14PTTS

Chapter 10, Problem 16PTTS

Chapter 10 Solutions

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

Ch. 10.1 - Prob. 1C Ch. 10.1 - Prob. 2C Ch. 10.1 - Prob. 3C Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E

Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29E Ch. 10.1 - Prob. 30E Ch. 10.1 - Prob. 31E Ch. 10.1 - Prob. 32E Ch. 10.2 - Prob. 1C Ch. 10.2 - Prob. 2C Ch. 10.2 - Prob. 3C Ch. 10.2 - Prob. 1E Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - Prob. 5E Ch. 10.2 - Prob. 6E Ch. 10.2 - Prob. 7E Ch. 10.2 - Prob. 8E Ch. 10.2 - Prob. 9E Ch. 10.2 - Prob. 10E Ch. 10.2 - Prob. 11E Ch. 10.2 - Prob. 12E Ch. 10.2 - Prob. 13E Ch. 10.2 - Prob. 14E Ch. 10.2 - Prob. 15E Ch. 10.2 - Prob. 16E Ch. 10.2 - Prob. 17E Ch. 10.2 - Prob. 18E Ch. 10.2 - Prob. 19E Ch. 10.2 - Prob. 20E Ch. 10.2 - 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Prob. 24E Ch. 10.8 - Prob. 25E Ch. 10.8 - Prob. 26E Ch. 10.8 - Prob. 27E Ch. 10.8 - Prob. 28E Ch. 10.8 - Prob. 29E Ch. 10.8 - Prob. 30E Ch. 10.8 - Prob. 31E Ch. 10.8 - Prob. 32E Ch. 10.8 - Prob. 33E Ch. 10.8 - Prob. 34E Ch. 10 - Prob. 1PSQ Ch. 10 - Prob. 2PSQ Ch. 10 - Prob. 3PSQ Ch. 10 - Prob. 4PSQ Ch. 10 - Prob. 5PSQ Ch. 10 - Prob. 6PSQ Ch. 10 - Prob. 7PSQ Ch. 10 - Prob. 8PSQ Ch. 10 - Prob. 9PSQ Ch. 10 - Prob. 10PSQ Ch. 10 - Prob. 11PSQ Ch. 10 - Prob. 1MCQ Ch. 10 - Prob. 2MCQ Ch. 10 - Prob. 3MCQ Ch. 10 - Prob. 4MCQ Ch. 10 - Prob. 5MCQ Ch. 10 - Prob. 6MCQ Ch. 10 - Prob. 7MCQ Ch. 10 - Prob. 8MCQ Ch. 10 - Prob. 9MCQ Ch. 10 - Prob. 10MCQ Ch. 10 - Prob. 11MCQ Ch. 10 - Prob. 1CR Ch. 10 - Prob. 2CR Ch. 10 - Prob. 3CR Ch. 10 - Prob. 4CR Ch. 10 - Prob. 5CR Ch. 10 - Prob. 6CR Ch. 10 - Prob. 7CR Ch. 10 - Prob. 8CR Ch. 10 - Prob. 9CR Ch. 10 - Prob. 10CR Ch. 10 - Prob. 11CR Ch. 10 - Prob. 12CR Ch. 10 - Prob. 13CR Ch. 10 - Prob. 14CR Ch. 10 - Prob. 15CR Ch. 10 - Prob. 16CR Ch. 10 - Prob. 17CR Ch. 10 - Prob. 18CR Ch. 10 - Prob. 19CR Ch. 10 - Prob. 20CR Ch. 10 - Prob. 21CR Ch. 10 - Prob. 22CR Ch. 10 - Prob. 23CR Ch. 10 - Prob. 24CR Ch. 10 - Prob. 25CR Ch. 10 - Prob. 26CR Ch. 10 - Prob. 1CT Ch. 10 - Prob. 2CT Ch. 10 - Prob. 3CT Ch. 10 - Prob. 4CT Ch. 10 - Prob. 5CT Ch. 10 - Prob. 6CT Ch. 10 - Prob. 7CT Ch. 10 - Prob. 8CT Ch. 10 - Prob. 9CT Ch. 10 - Prob. 10CT Ch. 10 - Prob. 11CT Ch. 10 - Prob. 12CT Ch. 10 - Prob. 13CT Ch. 10 - Prob. 14CT Ch. 10 - Prob. 15CT Ch. 10 - Prob. 1CST Ch. 10 - Prob. 2CST Ch. 10 - Prob. 3CST Ch. 10 - Prob. 4CST Ch. 10 - Prob. 5CST Ch. 10 - Prob. 6CST Ch. 10 - Prob. 7CST Ch. 10 - Prob. 8CST Ch. 10 - Prob. 9CST Ch. 10 - Prob. 10CST Ch. 10 - Prob. 11CST Ch. 10 - Prob. 12CST Ch. 10 - Prob. 2PTTS Ch. 10 - Prob. 3PTTS Ch. 10 - Prob. 4PTTS Ch. 10 - Prob. 5PTTS Ch. 10 - Prob. 6PTTS Ch. 10 - Prob. 7PTTS Ch. 10 - Prob. 8PTTS Ch. 10 - Prob. 9PTTS Ch. 10 - Prob. 10PTTS Ch. 10 - Prob. 11PTTS Ch. 10 - Prob. 12PTTS Ch. 10 - Prob. 13PTTS Ch. 10 - Prob. 14PTTS Ch. 10 - Prob. 15PTTS Ch. 10 - Prob. 16PTTS Ch. 10 - Prob. 1CP Ch. 10 - Prob. 2CP Ch. 10 - Prob. 3CP Ch. 10 - Prob. 4CP Ch. 10 - Prob. 5CP Ch. 10 - Prob. 6CP Ch. 10 - Prob. 7CP Ch. 10 - Prob. 8CP Ch. 10 - Prob. 9CP Ch. 10 - Prob. 10CP Ch. 10 - Prob. 11CP Ch. 10 - Prob. 12CP Ch. 10 - Prob. 13CP Ch. 10 - Prob. 14CP Ch. 10 - Prob. 16CP Ch. 10 - Prob. 17CP Ch. 10 - Prob. 18CP Ch. 10 - Prob. 19CP Ch. 10 - Prob. 20CP Ch. 10 - Prob. 21CP Ch. 10 - Prob. 22CP Ch. 10 - Prob. 23CP Ch. 10 - Prob. 25CP Ch. 10 - Prob. 26CP Ch. 10 - Prob. 27CP Ch. 10 - Prob. 28CP Ch. 10 - Prob. 29CP Ch. 10 - Prob. 30CP Ch. 10 - Prob. 31CP Ch. 10 - Prob. 32CP Ch. 10 - Prob. 33CP Ch. 10 - Prob. 34CP

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To find the equation of the line A B

Concept explainers

Videos

To find the equation of the line AB

Answer to Problem 15PTTS

Explanation of Solution

To find the equation of a line that is parallel to the line shown and passes through the point (−7,−6)

Answer to Problem 15PTTS

Explanation of Solution

To find the equation of a line that is perpendicular to the line shown and passes through the point (−3,4)

Answer to Problem 15PTTS

Explanation of Solution

To find the equation of a line that is perpendicular to the line shown and passes through the point (5,−2)

Answer to Problem 15PTTS

Explanation of Solution

To Graph : To graph the given lines

Explanation of Solution

Chapter 10 Solutions

Additional Math Textbook Solutions

To Graph :

To graph the given lines