please help me with these questions :(

Please ask only one question at a time.

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

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

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   =   $6\sqrt3$

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

Thank you! These were very helpful.