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1. Let \(O\) be the origin. Points \(P\) and \(Q\) lie in the first quadrant. The slope of line segment \(\overline{OP}\) is \(1\) and the slope of line segment \(\overline{OQ}\) is \(7.\) If \(OP = OQ,\) then compute the slope of line segment \(\overline{PQ}.\)

 

Note: The point \((x,y)\) lies in the first quadrant if both \(x\) and \(y\) are positive.

 

2. The line \(y = \frac{3x + 15}{4}\) intersects the circle \(x^2 + y^2 = 36\) at \(A\) and \(B\). Find the length of chord \(\overline{AB}\).

 

3. Let \(A = (4,-1)\)\(B = (6,2)\), and \(C = (-1,2)\). There exists a point \(X\) and a constant \(k\) such that for any point \(P\)\(PA^2 + PB^2 + PC^2 = 3PX^2 + k.\)
Find the constant \(k\).

 

Thank you for your help!

 

P.S. There were no images given for any of these problems.

 May 14, 2020
 #1
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Please ask only one question at a time.

 May 15, 2020
 #2
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Ok sorry. Is anyone still willing to help me out?

 May 15, 2020
 #3
avatar+8945 
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1.

 

P  is a point on the line   y = x   and   Q  is a point on the line  y = 7x.

 

So we can say...

 

Let  P  be the point  (a, a)           where  a > 0

Let  Q  be the point  (b, 7b)        where  b > 0

And  O  is the point  (0, 0)

 

Now we can make this equation:

 

the distance between  O  and  P   =   the distance between  O  and  Q

 

√[ a2 + a2 ]   =   √[ b2 + (7b)2 ]

                                                    Square both sides of the equation

a2 + a2   =   b2 + (7b)2

 

a2 + a2   =   b2 + 49b2

                                                    Combine like terms

2a2   =   50b2

                                                    Divide both sides by  2

a2   =   25b2

                                                    Take the positive square root of both sides

a   =   5b

 

the slope of  PQ   =   the slope between the points  (a, a)  and  (b, 7b)

 

the slope of  PQ   =   \(\dfrac{a-7b}{a-b}\)

                                                         Substitute  5b  in for  a

the slope of  PQ   =   \(\dfrac{5b-7b}{5b-b}\)

                                                         Combine like terms

the slope of  PQ   =   \(\dfrac{-2b}{4b}\)

                                                         Reduce the fraction by  2b

the slope of  PQ   =   \(-\dfrac{1}{2}\)

 

Here is a graph for this problem:  https://www.desmos.com/calculator/c9sjxvalqw

 May 15, 2020
 #4
avatar+8945 
0

2.

 

Let  (x, y)  be the solution to the system. We know it is true that

 

y  =  (3x + 15) / 4

 

We also know it is true that:

 

x2  +  y2   =   36

                                                          Since   y  =  \(\frac{3x+15}{4}\)  we can substitute  \(\frac{3x+15}{4}\)  in for  y

x2  +  \((\frac{3x+15}{4})^2\)   =   36

 

x2  +  \((\frac{3x+15}{4})(\frac{3x+15}{4})\)   =   36

 

x2  +  \(\frac{9x^2+90x+225}{16}\)   =   36

                                                          Multiply through by  16  to eliminate the denominator

16x2  +  9x2 + 90x + 225   =   576

                                                          Combine like terms

25x2  +  90x  +  225   =   576

                                                          Subtract  576  from both sides of the equation

25x2  +  90x  -  351   =   0

                                                         We can use the quadratic formula to solve for  x

x   =  \(\dfrac{-9\pm12\sqrt3}{5}\)

 

We can use the equation  y  =  (3x + 15) / 4   to find the y-coordinates of the intersection points.

 

When   x  =  \(\frac{-9\ +\ 12\sqrt3}{5}\)   ,   y  =  \(\frac14\cdot(\ 3(\ \frac{-9\ +\ 12\sqrt3}{5}\ )+15\ )\ =\ \frac{12\ +\ 9\sqrt3}{5}\)

 

When   x  =  \(\frac{-9\ -\ 12\sqrt3}{5}\)   ,   y  =  \(\frac14\cdot(\ 3(\ \frac{-9\ -\ 12\sqrt3}{5}\ )+15\ )\ =\ \frac{12\ -\ 9\sqrt3}{5}\)

 

So the intersection points are:

 

\(\big(\frac{-9\ +\ 12\sqrt3}{5},\ \frac{12\ +\ 9\sqrt3}{5}\big)\)     and     \(\big(\frac{-9\ -\ 12\sqrt3}{5},\ \frac{12\ -\ 9\sqrt3}{5}\big)\)

 

Check:  https://www.desmos.com/calculator/jfndjchrhd

 

We can find the distance between those two points using the distance formula.

 

distance   =   \(\sqrt{\Big(\frac{-9\ +\ 12\sqrt3}{5}\ -\ \frac{-9\ -\ 12\sqrt3}{5}\Big)^2\ +\ \Big(\frac{12\ +\ 9\sqrt3}{5}\ -\ \frac{12\ -\ 9\sqrt3}{5}\Big)^2}\)

 

distance   =   \(\sqrt{\Big(\frac{24\sqrt3}{5}\Big)^2\ +\ \Big(\frac{18\sqrt3}{5}\Big)^2}\)

 

distance   =   \(\sqrt{\frac{1728}{25}\ +\ \frac{972}{25}}\)

 

distance   =   \(\sqrt{\frac{2700}{25}}\)

 

distance   =   \(6\sqrt3\)

.
 May 15, 2020
 #5
avatar+8945 
0

3.

 

Let  P  be the point  (x, y)

 

So...

 

\(PA\ =\ \sqrt{(x-4)^2+(y+1)^2} \\~\\ PB\ =\ \sqrt{(x-6)^2+(y-2)^2} \\~\\ PC\ =\ \sqrt{(x+1)^2+(y-2)^2} \\~\\~\\ PA^2+PB^2+PC^2\ =\ (x-4)^2+(y+1)^2\ +\ (x-6)^2+(y-2)^2\ +\ (x+1)^2+(y-2)^2 \\~\\ PA^2+PB^2+PC^2\ =\ 3 x^2 - 18 x + 3 y^2 - 6 y + 62\\~\\ PA^2+PB^2+PC^2\ =\ 3( x^2 - 6 x + y^2 - 2 y) + 62\\~\\ PA^2+PB^2+PC^2\ =\ 3( x^2 - 6 x +9-9\ +\ y^2 - 2 y+1-1) + 62\\~\\ PA^2+PB^2+PC^2\ =\ 3( (x-3)^2-9\ +\ (y-1)^2-1) + 62\\~\\ PA^2+PB^2+PC^2\ =\ 3( (x-3)^2\ +\ (y-1)^2-10) + 62\\~\\ PA^2+PB^2+PC^2\ =\ 3( (x-3)^2\ +\ (y-1)^2)-30 + 62\\~\\ PA^2+PB^2+PC^2\ =\ 3( (x-3)^2\ +\ (y-1)^2)+32\)

 

 

So  X  must be the point  (3, 1)   and  k  must be  32

 May 15, 2020
 #6
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+1

Thank you! These were very helpful.

 May 15, 2020

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