We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

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

0
288
2
+129

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

### 2+0 Answers

#1
+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

Jan 28, 2018
#2
+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:

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

Jan 29, 2018