Find all points on the x-axis that are a distance 6 from P(7, -5).

Question

I'm very confused on how to go about solving this. I ended with x^2-14x+38=0 and that doesn't factor so I'm stuck. Thanks for any help you can provide.

Answer 1

The locus,or path , marked out by a point which moves such that its distance from a fixed point is constant is a circle with the fixed point as the center and the distance as the radius. Thus the points which are 6 units from the point (7, -5) is a circle centered at (7, -5) with a radius of 6 units.

The standard form of the equation of a circle with center at (h, k) and radius r is :         (x - h)²  +  (y - k)²   =  r ²  

 

The equation of the circle formed by the points which are 6 units from the point (7, -5) is obtained by substituting

h= 7, k = -5; and r = 6 into the standard form of the equation of a circle.

 

This gives the equation       (x -7)²   + (y - -5)² = 6 ²  

 

 

                                         =         (x -7)² + (y + 5)² = 6 ²                               

 

The x axis is the line y = 0, so the points of intersection would be all the points that satisfy both equations. We find these points by solving the system of equations formed by the equation of the circle and the line y=0.

 

 

Substituting y =0 into the equation   (x -7)² + (y +  5)² = 6 ² gives     (x -7)² + (0 + 5)² = 6 ² 

 

simplifying we get  (x -7)² + ( 5)² = 6 ² 

 

                            (x -7)²   +25     =      36

                                          -25           - 25

 

                                (x -7)²             =    11

 

Taking the square root of both sides, we get  x -7 = + root 11 or - root 11.

Therefore the points of intersection are x = 7 + root 11 and   7 - root 11.

  

Answer 2

If you look at it graphically there should be two points a distance 6 from P(7, -5).

 

What formula one needs is the distance between two points P1(x₁ , y₁) and P2(x₂ , y₂).

 

|P1,P2| = ((x₂ - x₁)² + (y₂ - y₁)²)^.5

 

Let P1 = P(7, -5).

 

P2 = P( x₂, 0)  y₂ = 0 since the points are on the x-axis.

 

Now plug everything into the distance formula.

 

6 = ((x₂ - 7)² + (0 + 5)²)^.5

 

squaring both sides

 

36 = (x₂ - 7)² + 25

 

subtracting 25 from both sides.

 

11 = (x₂ - 7)²

 

taking the square root of both sides. (Remember: Taking the square root gives two answers; a positive and a negative one.

 

11^.5 = x₂ - 7  and 11^.5 = -x₂ + 7

 

So the two points are: (11^.5 + 7, 0)  and (-11^.5 + 7, 0) 

Answer 3

The x-axis has points that strictly have a y-coordinate of zero.  Based on this, there will be two points on the x-axis that is a distance of 6 from the point (7, -5).

 

We can use the distance formula to find those points.

 

d = √[(x₁ - x₂)² + (y₁ - y₂)²]

 

where:

d = 6

Point 1 = (7, -5)

Point 2 = (x₂ , 0)

Point 3 = (x₃ , 0)

 

Point 2 and Point 3 are the points on the x-axis. 

 

6 = √[(7 - x₂)² + (-5 - 0)²]

 

 

Square both sides of the equation.

 

36 = (7 - x₂)² + (-5)²

 

 

Solve for x₂ from this equation. 

 

36 = (7 - x₂)² + 25

 

11 = (7 - x₂)²

 

 

Square-root both sides of the equation.

 

±√11 = 7 - x₂

 

 

Subtract 7 on both sides of the equation.

 

-7 ± √11 = -x

 

 

Because of the plus/minus sign, there will be two x values.  Onc for x₂ and the other for x₃.

 

-7 - √11 = -x₂            and             -7 + √11 = -x₃

 

 

Divide both sides of both equations by -1.

 

7 + √11 = x₂                 and              7 - √11 = x₃

 

 

Your two points are then

 

(7 + √11 , 0)      and       (7 - √11 , 0)

Answer 4

Hello Sky,

 

This is an exercise in using the equation to find the distance between two points in your case the first point is  P(7, -5) and the other call it Q is on the x-axis so it is  Q(X,0) not y is zero there. The distance between P and Q is 6

 

d² =(X₂ - X₁)² + (Y₂ - Y₁)²

 

6² = (X-7)² +(0 -(-5))²

 

36 = (X-7)² + 25

 

9 = (X-7)²   

 

(X-7) = ± 3  The possible points are  X = 10 and X = 4

Answer 5

Your problem involved the Pythagorean theorem.

 

Plot point P and draw in the segment from P to (7, 0) on the x-axis.  This segment is the altitude of an isosceles triangle.  The sides are each length 6.  You need the coordinates of the base vertices.

 

Use the Pythagorean to find the missing side.

 

a² + b² = r², where a = 5 and r = 6

 

b² = 36 - 25 = 11

 

b = sqrt(11) = + 3.32

 

 

Now for the x-coordinates of the base vertices.  They are equidistant from (7, 0) at (7 + 3.32, 0).

 

One is at (7 - 3.32, 0) or (3.68, 0).

 

The other is at (7 + 3.32, 0) or (10.32, 0).