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 #3
avatar+9142 
0
Sep 24, 2020
 #1
avatar+9142 
+1

The sum of the measures of all the arcs in a circle is  360°

 

So... looking at the circle in problem #9....we can make the following equation:

 

61°   +   (5x - 7)°   +   34°   +   (9x - 22)°   =   360°

                                                                                Remove the degree sign from every term.

61   +   5x - 7   +   34   +   9x - 22   =   360

                                                                                Combine like terms.

14x   +  66   =  360

                                   Subtract  66  from both sides of the equation.

14x   =   294

                                   Divide both sides of the equation by  14

x   =   21

 

Now that we know what  x  is, we can find the measures of all the arcs.

 

\(\begin{array}{ccccccc} m\stackrel{\large\frown}{GK}& =& (9x-22)^{\circ}& =&(9(21)-22)^\circ&=&167^\circ\\~\\ m\stackrel{\large\frown}{HJ}& =& (5x-7)^{\circ}& =&(5(21)-7)^\circ&=&98^\circ\\~\\ m\stackrel{\large\frown}{HGJ}& =& m\stackrel{\large\frown}{HG}+m\stackrel{\large\frown}{GK}+m\stackrel{\large\frown}{KJ} & =& 61^\circ+167^\circ+34^\circ&=&262^\circ\\~\\ m\stackrel{\large\frown}{GKJ}&=&m\stackrel{\large\frown}{GK}+m\stackrel{\large\frown}{KJ} &=&167^\circ+34^\circ&=&201^\circ \end{array}\)

.
Sep 24, 2020
 #1
avatar+9142 
+2

We are given:

 

A  =  (4, 1)

B  =  (6, 2)

C  =  (-1, 2)

 

And we can let:

 

P  =  (x, y)

 

By the Pythagorean Theorem/distance formula we can say:

 

PA2  =   (x - 4)2  +  (y - 1)2

 

PB2  =   (x - 6)2  +  (y - 2)2

 

PC2  =   (x + 1)2  +  (y - 2)2

 

Then....

 

PA2 + PB2 + PC2   =   (x - 4)2  +  (y - 1)2   +   (x - 6)2  +  (y - 2)2   +   (x + 1)2  +  (y - 2)2

 

Expand each term, then combine like terms to get:

 

PA2 + PB2 + PC2   =   3x2  -  18x  +  3y2  -  10y  +  62

 

(I used this to do that because I am lazy.)

 

Next we want to get the expression on the right side of the equation into the form:

 

3PQ2  +  k

 

which is:

 

3[ (x - something)2  +  (y - something)2 ] + k

 

To do that, we need to complete the squares of the x terms and the y terms.

 

PA2 + PB2 + PC2   =   3(x2  -  6x)  +  3(y2  -  \(\frac{10}{3}\)y)  +  62

 

PA2 + PB2 + PC2   =   3(x2  -  6x + 9 - 9)  +  3(y2  -  \(\frac{10}{3}\)y + \(\frac{25}{9}\)  -  \(\frac{25}{9}\))  +  62

 

PA2 + PB2 + PC2   =   3( (x - 3)2 - 9)  +  3( (y - \(\frac53\))2  -  \(\frac{25}{9}\))  +  62

 

PA2 + PB2 + PC2   =   3(x - 3)2 - 27  +  3(y - \(\frac53\))2  -  \(\frac{25}{3}\)  +  62

 

PA2 + PB2 + PC2   =   3(x - 3)2  +  3(y - \(\frac53\))2  -  \(\frac{25}{3}\)  +  62 - 27

 

PA2 + PB2 + PC2   =   3[ (x - 3)2  +  (y - \(\frac53\))2 ]   -  \(\frac{25}{3}\)  +  62 - 27

 

PA2 + PB2 + PC2   =   3[ (x - 3)2  +  (y - \(\frac53\))2 ]   +   \(\frac{80}{3}\)

 

(Check)

 

Now it is in the desired form, and so we can pick out that    k   =   \(\frac{80}{3}\)

 

BTW, I came across this answer. It works out a bit more cleanly if  A  =  (4, -1).  Just wanted to mention this in case that is what you meant. smiley

Sep 22, 2020