Could someone please show me how to solve these?

Question

For the first problem, we have "Point P is on the x-axis, 13 units from point C(7, 5). What are the possible measures of the angles?"

Second, I have "What are the coordinates of the point P on the x-axis such that the slope
of a line from P to A(1, 4) is twice the slope of a line from P to B(9, 10)?"

Oops. I miscopied the first problem.
It should read as "Point P is on the x-axis, 13 units from point C(7, 5).
What are the possible coordinates of P?"
Sorry for the inconvenience.

Answer 1

P is at the intersections of the x axis and a circle of radius 13 that is centered at (7,5).
http://i233.photobucket.com/albums/ee195/DWRead/munch_simmons2.jpg

PC forms the hypotenuse of a right triangle. Using the Pythagorean theorem, the base of the triangle is 12. Thus, P=(7±12,0)
----------------------------------
A=(1,4), B=(9,10), and let P = (x,0).
slope of PA = (4-0)/(1-x) = 4/(1-x)
slope of PB = (10-0)/(9-x) = 10/(9-x)

slope of PA = 2*slope of PB
4/(1-x) = 2*10/(9-x)
Cross-multiply and solve for x:
4(9-x) = 20(1-x)
36-4x = 20-20x
16x = -16
x = -1
P = (-1,0)
http://i233.photobucket.com/albums/ee195/DWRead/munch_simmons3.jpg

Answer 2

a)
Given:
C(7,5)
P(x, 0) since P is on the x-axis, y = 0
13 units = distance between P and C

Find: Possible measure of the angles? (I really don't know what angles are you talking about here. But I assume it's the angle that the line segment connecting P and C makes with the x-axis.)

Well, P could be on the "left" side of point C or on the "right" side of point C.

When P is on the right side of point C:

Dpx = √[(Xp - Xc)^2 + (Yp - Yc)^2]
13 = √[(x - 7)^2 + (0 - 5)^2]
(13)^2 = {√[(x-7)^2 + (-5)^2]}^2
169 = (x - 7)^2 + 25
(x - 7)^2 + 25 = 169
(x - 7)^2 = 169 - 25
(x - 7)^2 = 144
√[(x - 7)^2] = √144
x - 7 = 12
x = 12 + 7
x = 19

So, P(19, 0) and C(7, 5)
θ = arctan[(0 - 5)/(19 - 7)]
θ = 157.4° ANS

When P is on the left side of C:
Dpx = √[(Xc - Xp)^2 + (Yc - Yp)^2]
13 = √[(7 - x)^2 + (5 - 0)^2]
(13)^2 = {√[(7 - x)^2 + (5)^2]}^2
169 = (7 - x)^2 + 25
(7 - x)^2 + 25 = 169
(7 - x)^2 = 169 - 25
(7 - x)^2 = 144
√[(7 - x)^2] = √144
7 - x = 12
- x = 12 - 7
x = - 12 + 7
x = - 5
So P(-5,0) and C(7,5)
θ = arctan[(5 - 0)/(7 - -5)
θ = 22.6° ANS

b) Given:
A(1, 4), B(9, 10)
Mpa slope from P to A is twice Mpb the slope from P to B
or Mpa = 2Mpb

Find:
P(Xp, Yp), coordinates of P

Solution:
Mpa = (Ya - Yp)/(Xa - Xp)
Mpa = (4 - Yp)/(1 - Xp)

Mpb = (Yb - Yp)/(Xb - Xp)
Mpb = (10 - Yp)/(9 - Xp)

But, from the given:
Mpa = 2Mpb
(4 - Yp)/(1 - Xp) = 2[(10 - Yp)/(9 - Xp)]

Since point P is on the x-axis, then Yp = 0. Hence,
(4 - 0)/(1 - Xp) = 2[(10 - 0)/(9 - Xp)]
4/(1 - Xp) = 2[10/(9 - Xp)]
2/(1 - Xp) = 10/(9 - Xp)
1/(1- Xp) = 5/(9 - Xp)
(1 - Xp)(9 - Xp)[1/(1 - Xp)] = (1 - Xp)(9 - Xp)[5/(9 - Xp)]
9 - Xp = 5(1 - Xp)
9 - Xp = 5 - 5Xp
- Xp + 5Xp = 5 - 9
4Xp = - 4
Xp = -4/4
Xp = -1
Therefore, P(-1, 0) ANS

teddy boy

Answer 3

1)P(20,0) or P(-6, 0) Edit: Not Correct! 2 Gents below have the correct Answer P(19,0) or P(-5,0) 1 "Thumb Up" for both.

2) y-y1 = m(x-x1)

m1 = 0-4/x-1 = -4/x-1

m2 = 0-10/x-9 = -10/x-1

m1 = 2m2 = -4/x-1 = -20/x-9 =>> x-9 = 5x-5 =
4x = -4 = x = -1

P(-1,0)

Best Regards.

Answer 4

Sorry Nope can't help

Answer 5

Can you show the table so that I can comprehence.

Answer 6

i'd need to see the table....sorry dude but goodluck!

Answer 7

need to have a table!