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point A is located at (1,4). Point P ar (3,5) is 1/3 the distance from A to point B.What are the coordinates of point b (SHOW WORK)

 Jan 29, 2018
 #1
avatar+111331 
+2

Point A is located at (1,4). Point P at (3,5) is 1/3 the distance from A to point B.What are the coordinates of point b

 

Let m be the x coordinate of B

 

So

 

x coordinate of A   +  3 times ( x coordinate of P - x coordinate of A)  =  m

 

1   +  3 ( 3 - 1)  =  m

1  + 3(2)  = m

1 + 6  = m

7  = m

 

Similarly

 

Let n be the y coordinate of B

 

So

 

y coordinate of A   +  3 times ( y coordinate of P - y coordinate of A)  =  n

 

4  +  3 ( 5 - 4)  =  n

4  + 3(1)  = m

4 + 3  = m

7  = m

 

 

So B =  (m, n)  =  (7, 7)

 

cool cool cool

 Jan 29, 2018
 #2
avatar+24995 
+2

point A is located at (1,4). Point P ar (3,5) is 1/3 the distance from A to point B.

What are the coordinates of point b (SHOW WORK)

 

Formula:

\(\begin{array}{|l|rcll|} \hline & \mathbf{(1-\lambda)\vec{A} + \lambda\vec{B}} &\mathbf{=}& \mathbf{\vec{P}} \\ \hline \lambda = 0 & (1-0)\vec{A} + 0\cdot \vec{B} &=& \vec{A} \ \checkmark \\\\ \lambda = 1 & (1-1)\vec{A} + 1\cdot \vec{B} &=& \vec{B} \ \checkmark \\\\ \lambda = \dfrac13 & \left(1-\dfrac13 \right)\vec{A} + \dfrac13\cdot \vec{B} &=& \vec{P} \quad & | \quad \vec{B} =\ ? \\ \hline \end{array}\)

 

\(\vec{B} =\ ?\)

\(\begin{array}{|rcll|} \hline \left(1-\dfrac13 \right)\vec{A} + \dfrac13\cdot \vec{B} &=& \vec{P} \quad & | \quad \vec{B} =\ ? \\\\ \dfrac23 \vec{A} + \dfrac13 \vec{B} &=& \vec{P} \quad & | \quad \cdot 3 \\ \\ 2 \vec{A} + \vec{B} &=& 3 \vec{P} \quad & | \quad - 2 \vec{A} \\\\ \mathbf{\vec{B}} &\mathbf{=}& \mathbf{3 \vec{P} - 2 \vec{A}}\quad & | \quad \vec{A} = \binom{1}{4} \quad \vec{P} = \binom{3}{5} \\\\ \vec{B} &=& 3\binom{3}{5} - 2 \binom{1}{4} \\\\ \vec{B} &=& \binom{9}{15} - \binom{2}{8} \\\\ \vec{B} &=& \binom{9-2}{15-8} \\\\ \mathbf{\vec{B}} &\mathbf{=}& \mathbf{\binom{7}{7}} \\ \hline \end{array}\)

 

 

laugh

 Jan 29, 2018

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