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Blababa [14]
3 years ago
15

(47 Points)

Mathematics
1 answer:
Semenov [28]3 years ago
4 0

case 1)

Line 1

Let A (2,5) B (-3,-5)

Line 2

Let C (3,0) D (0,-3)

Find the equation of the line 1

Line 1

Let A (2,5) B (-3,-5)

slope m=(y2-y1) /(x2-x1)------> m=(-5-5) /(-3-2)-----> m=2

with m=2 and the point A (2,5) find the equation of the line 1

y-y1=m*(x-x1)------->y-5=2*(x-2)------>y=2x-4+5------> y=2x+1

Find the equation of the line 2

Line 2

Let C (3,0) D (0,-3)

slope m=(y2-y1) /(x2-x1)------> m=(-3-0) /(0-3)-----> m=1

with m=1 and the point C (3,0) find the equation of the line 2

y-y1=m*(x-x1)------->y-0=1*(x-3)------>y=x-3

using a graph tool

see the attached figure N 1

the solution is the point (-4,-7)

case 2)

Line 1

Let A (1,1) B (2,3)

Line 2

Let C (0,3) D (2,5)

Find the equation of the line 1

Line 1

Let A (1,1) B (2,3)

slope m=(y2-y1) /(x2-x1)------> m=(3-1) /(2-1)-----> m=2

with m=2 and the point A (1,1) find the equation of the line 1

y-y1=m*(x-x1)------->y-1=2*(x-1)------>y=2x--2+1------> y=2x-1

Find the equation of the line 2

Line 2

Let C (0,3) D (2,5)

slope m=(y2-y1) /(x2-x1)------> m=(5-3) /(2-0)-----> m=1

with m=1 and the point C (0,3) find the equation of the line 2

y-y1=m*(x-x1)------->y-3=1*(x-0)------>y=x+3

using a graph tool

see the attached figure N 2

the solution is the point (4,7)

case 3)

Line 1

Let A (1,0) B (0,-1)

Line 2

Let C (0,3) D (-2,-1)

Find the equation of the line 1

Line 1

Let A (1,0) B (0,-1)

slope m=(y2-y1) /(x2-x1)------> m=(-1-0) /(0-1)-----> m=1

with m=1 and the point A (1,0) find the equation of the line 1

y-y1=m*(x-x1)------->y-0=1*(x-1)------>y=x-1

Find the equation of the line 2

Line 2

Let C (0,3) D (-2,-1)

slope m=(y2-y1) /(x2-x1)------> m=(-1-3) /(-2-0)-----> m=2

with m=2 and the point C (0,3) find the equation of the line 2

y-y1=m*(x-x1)------->y-3=2*(x-0)------>y=2x+3

using a graph tool

see the attached figure N 3

the solution is the point (-4,-5)


case 4)

Line 1

Let A (2,0) B (0,-2)

Line 2

Let C (4,5) D (3,3)

Find the equation of the line 1

Line 1

Let A (2,0) B (0,-2)

slope m=(y2-y1) /(x2-x1)------> m=(-2-0) /(0-2)-----> m=1

with m=1 and the point A (2,0) find the equation of the line 1

y-y1=m*(x-x1)------->y-0=1*(x-2)------>y=x-2

Find the equation of the line 2

Line 2

Let C (4,5) D (3,3)

slope m=(y2-y1) /(x2-x1)------> m=(3-5) /(3-4)-----> m=2

with m=1 and the point C (4,5) find the equation of the line 2

y-y1=m*(x-x1)------->y-5=2*(x-4)------>y=2x-8+5-----> y=2x-3

using a graph tool

see the attached figure N 4

the solution is the point (1,-1)

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Complete Question

The complete question is shown on the first uploaded image

Answer:

First Question

Option A is correct

Second  Question

Option C is correct

Third   Question

     D =  A^{-1}  *  B^{-1} *  C^{-1}

Fourth   Question

  So substituting for D in  (ABC) D =  I

                 (ABC) *  A^{-1}  *  B^{-1} *  C^{-1} =  I

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This proof that  ABC is invertible

Step-by-step explanation:

From the question we are told that

   A , B and  C are invertible which means that A^{-1} , B^{-1}, C^{-1} exist

Now

 From the question

          (ABC) D =  I

Where I is an identity matrix

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          A^{-1}  A BCD =  A^{-1} * I

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Now when we multiply both sides by  B^{-1}  we have  

         B^{-1 } *I BCD =  A^{-1}  *  B^{-1}

         I CD =  A^{-1}  *  B^{-1}

Now when we multiply both sides by  C^{-1}  we have  

          C^{-1} * I CD =  A^{-1}  *  B^{-1} *  C^{-1}

              I D =  A^{-1}  *  B^{-1} *  C^{-1}

                 D =  A^{-1}  *  B^{-1} *  C^{-1}

So substituting for D in the above equation

                 (ABC) *  A^{-1}  *  B^{-1} *  C^{-1} =  I

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Step-by-step explanation:

A. y squared minus y squared z minus 10 y z minus 5

B. y squared minus y squared z minus 5 y z minus 5

C. 7 y squared minus y squared z minus 10 y z minus 5

D. 7 y squared minus y squared z minus 2 y z minus 5

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3y^2 - 6yz - 7 + 4y^2 - 4yz + 2 - y^2z

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