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Coordinates plane H(-5,-6) & J(5,-1)

.Find the Point Q that is 4/5 the distance from J to H

show work

 Jan 28, 2018
 #1
avatar+101085 
+1

 

IMHO, It's easier to figure point Q if we realize that it also must be equal to 1/5 of the distance from H to J

 

So....we can find point Q as follows :

 

Q  =   

 

[ -5  +  (1/5) ( 5 - - 5) ,  -6 + (1/5) ( -1 - -6)  ]  =

 

[ -5 + (1/5) (10)  , - 6 + (1/5) (5) ]  =

 

[ - 5 + 2, - 6 + 1 ]

 

{ -3, - 5)  =  Q

 

 

cool cool cool

 Jan 28, 2018
 #2
avatar+22260 
+2

Coordinates plane H(-5,-6) & J(5,-1).
Find the Point Q that is 4/5 the distance from J to H

 

Formula:

\(\begin{array}{|rcll|} \hline \mathbf{\vec{Q}} &\mathbf{=}& \mathbf{(1-\lambda)\vec{J} + \lambda\vec{H}} \quad & | \quad\lambda = \dfrac45 \\\\ \vec{Q} &=& \left(1-\dfrac45 \right)\vec{J} + \dfrac45\cdot \vec{H} \\\\ &=& \dfrac15 \vec{J} + \dfrac45 \vec{H} \quad & | \quad \vec{J} = \dbinom{5}{-1} \quad \vec{H} = \dbinom{-5}{-6} \\\\ &=& \dfrac15 \dbinom{5}{-1} + \dfrac45 \dbinom{-5}{-6} \\\\ &=& \dbinom{1}{-\dfrac15} + \dbinom{-4}{-\dfrac{24}{5}} \\\\ &=& \dbinom{1}{-\dfrac15} - \dbinom{4}{\dfrac{24}{5}} \\\\ &=& \dbinom{1-4}{-\dfrac15-\dfrac{24}{5 }} \\\\ &=& \dbinom{-3}{-\dfrac{25}{5 }} \\\\ \mathbf{\vec{Q}} & \mathbf{=}& \mathbf{\dbinom{-3}{-5}} \\ \hline \end{array}\)

 

laugh

 Jan 29, 2018

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