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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 $$k$$.

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

May 14, 2020

#1
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May 15, 2020
#2
0

Ok sorry. Is anyone still willing to help me out?

May 15, 2020
#3
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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
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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     $$\big(\frac{-9\ -\ 12\sqrt3}{5},\ \frac{12\ -\ 9\sqrt3}{5}\big)$$

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
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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
+1

Thank you! These were very helpful.

May 15, 2020