+0

# graph question

0
83
6
+187

Feb 22, 2020

#1
+531
+3

Q  (16, 0)

T   (8, 0)

W  (10, 12)

P   (12, 24)       ( P is 'somewhere over the rainbow' )

R   (6, 12)

S   (14, 12)

Feb 22, 2020
edited by Dragan  Feb 22, 2020
#2
+108679
+2

A midpoint of 2 points is the (average value of the x's, average value of the y's)

T is on the x axis (y=0)

U and V are both on the line y=6,

So R, W and S must be on the line  y=12

And P must have a y value of 24.

Mmm

Here is a graphical solution.  (It is not the logic solution that you need though.)

It appears that Dragan has beaten me to it.   Nice work Dragan.

But neither of us has provided an answer with explained logic.

Maybe Dragan does have one.  Do you Dragan?

Feb 22, 2020
#3
+531
+1

Yours too, Melody!

Dragan  Feb 22, 2020
edited by Dragan  Feb 22, 2020
#4
+108679
+1

Thanks Dragan,

I expect yours is better because I let Geogebra do half the work for me.

I have not worked out how to do it without Geogebra helping.

Would you like to put up your logic?

Melody  Feb 22, 2020
#5
+24388
+2

In the diagram, O is lacated at (0,0), U is located at (7,6),
and V is located at (11,6).
Assume that R, S, T, U, V, and W are midpoints.

$$1)\\ \begin{array}{|rcl|rcl|rcl|} \hline R &=& U + \dfrac{TR}{2} & \dfrac{TR}{2}+T &=& U & T &=& \dfrac{Q}{2} \\ && & \dfrac{TR}{2}+\dfrac{Q}{2} &=& U \\ && & \dfrac{TR}{2} &=& U-\dfrac{Q}{2} \\ R &=& U + U-\dfrac{Q}{2} \\ \mathbf{R} &=& \mathbf{2U-\dfrac{Q}{2}} \\ \hline S &=& T+TS & T+\dfrac{TS}{2} &=& V & T &=& \dfrac{Q}{2} \\ & & & \dfrac{Q}{2}+\dfrac{TS}{2} &=& V \quad | \quad *2 \\ & & & Q+TS &=& 2V \\ & & & TS &=& 2V-Q \\ S &=& \dfrac{Q}{2} + 2V-Q \\ \mathbf{S} &=& \mathbf{2V-\dfrac{Q}{2}} \\ \hline \end{array}$$

$$2)\\ \begin{array}{|rcl|rcl|rcl|} \hline && &P &=& S + \dfrac{QP}{2} & QP &=& 2(S-Q) \\ && &P &=& S + (S-Q) \\ && &P &=& 2S-Q & S&=&2V-\dfrac{Q}{2} \\ && &P &=& 4V-Q-Q \\ && &\mathbf{P} &=& \mathbf{4V-2Q} \\ \hline && &P &=& 2R & R&=&2U-\dfrac{Q}{2} \\ && &\mathbf{P} &=& \mathbf{4U-Q} \\ P= 4U-Q &=& 4V-2Q \\ 4U-Q &=& 4V-2Q \\ \mathbf{ Q }&=& \mathbf{4V-4U} \\ \hline \end{array} \\ 3)\\ \begin{array}{|rcl|rcl|rcl|} \hline W&=& \dfrac{1}{2}( R+S )& R&=&2U-\dfrac{Q}{2} & S&=&2V-\dfrac{Q}{2} \\ W&=& \dfrac{1}{2}( 2U-\dfrac{Q}{2}+2V-\dfrac{Q}{2} ) \\ W&=& \dfrac{1}{2}( 2U+2V-Q ) \\ \mathbf{ W }&=& \mathbf{U+V-\dfrac{Q}{2}} \\ \hline \end{array}$$

$$\text{Summary:}\quad Q=4V-4U \\ \begin{array}{|rcll|} \hline T&=& \dfrac{Q}{2} \\ R&=& 2U-\dfrac{Q}{2} \\ S&=& 2V-\dfrac{Q}{2} \\ W&=& U+V-\dfrac{Q}{2} \\ P&=& 4U-Q \\ \hline \end{array}$$

Substitute $$Q$$:

$$\begin{array}{|rcll|} \hline T &=& -2U+2V \\ Q &=& -4U+4V \\ R &=& 4U-2V \\ S &=& 2U \\ W &=& 3U-V \\ P &=& 8U-4V \\ \hline \end{array}$$

Substitute $$U(7,6)$$ and $$V(11,6)$$:

$$\begin{array}{|rcll|} \hline \mathbf{T} &=& -2(7,6)+2(11,6) = (-14+22, -12+12) =\mathbf{(8,0)}\\ \mathbf{Q} &=& -4(7,6)+4(11,6) = (-28+44, -24+24) =\mathbf{(16,0)}\\ \mathbf{R} &=& 4(7,6)-2(11,6) = (28-22, 24-12) =\mathbf{(6,12)}\\ \mathbf{S} &=& 2(7,6) = \mathbf{(14, 12)} \\ \mathbf{W} &=& 3(7,6)-(11,6) = (21-11, 18-6) =\mathbf{(10,12)}\\ \mathbf{P} &=& 8(7,6)-4(11,6) = (56-44, 48-24) =\mathbf{(12,24)}\\ \hline \end{array}$$

Feb 22, 2020
edited by heureka  Feb 23, 2020
#6
+108679
+2

Thanks Heureka :)

Melody  Feb 22, 2020