Question

Find the coordinates of a point P on the linesegment joining A(1, 2) and B(6, 7) such that AP= 2/5 AB

CORRECT ANSWER WILL GET EXTRA POINTS AND BRAINLIST ANSWER



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Answer 1

Answer:

Consider the problem

Consider the problemAP=

Consider the problemAP= 5

Consider the problemAP= 52

Consider the problemAP= 52

Consider the problemAP= 52 AB

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PB

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP =

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 3

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=(

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+3

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 ,

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+3

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=(

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 5

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515 ,

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515 , 5

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515 , 520

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515 , 520

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515 , 520 )

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515 , 520 )=(3,4)

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515 , 520 )=(3,4)

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515 , 520 )=(3,4)

Consider the problemAP= 52 AB⇒5AP=2(AP+PB)⇒5AP=2AP+2PB⇒3AP=2PB⇒ PBAP = 32 ⇒AP:PB=2:3 And(x,y)=( 2+32×6+3×1 , 2+32×7+3×2 )=( 515 , 520 )=(3,4) So, coordinates of point P is (3,4)

Step-by-step explanation:

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Answer 2

Answer:

your answer is right bro