Equidistant from (-3,0) and (1,4) and at a distance 5 from (-1,7)

Question

I Can't solve this problem

Answer 1

This problem, as you indicated, uses the Distance Formula.  It is based on the fact that the distance between two points in the x-y coordinate system is the hypotenuse of the right triangle that has a horizontal side between the two x values and a vertical side between the y values.  The distance formula is:

       d =  SQRT  ( (x₂ – x₁)² + (y₂ – y₁)² )

The problem states that a point is equidistant from (-3,0) and (1,4).  There are a whole line of such points (and if you draw the line segment between the two given points, bisect it, and draw the perpendicular line, that’s the one we’re talking about).

The problem also says “distance 5 from (-1,7).”  That means that the circle with center (-1,7) and radius 5 goes through that point.  But, we know that this circle might cross this line at two, one, or none at all points.  So, we must be careful.

First, the line with points (x,y) equidistant from (-3,0) and (1,4).  Set the distances equal:

     d = SQRT ( (x - (-3))^2 + (y - 0)^2 ) )  =  SQRT (x – 1)^2 + (y-4)^2 )
                (x+3)^2 + y^2 = (x-1)^2 + (y-4)^2                     [squaring both sides]
                x^2 + 6x + 9 + y^2  =  x^2 – 2x + 1 + y^2 – 8x + 16              [expanding]
                        8y = -8x + 8           [collecting terms and cancelling]
                         y = -x + 1              [dividing by 8]

Now, for a point (x,y) on this line to be a distance 5 from point (-1,7), the distance formula says:
       d = 5  =  SQRT ( (x - (-1))^2 + ( y -7 )^2 )
              25  = x^2 + 2x +1  + y^2 -14y + 49                  [squaring both sides]
              25 = x^2 + 2x + 1   + (-x+1)^2 - 14(-x + 1) + 49     [replacing y with  (-x+1)]
               0 = x^2 + 7x + 6                     [collect terms and divide by 2]
               0 = (x+1)(x+6)                    [either factor or use quadratic formula]
      So,   x=-1  or  x=-6

And, putting those values into the formula for the line y = -x + 1:
       We have the points  (-1,2) and (-6,7)

Checking (very important):
  Is the distance from (-3,0) to (-1,2) equal to the distance from (1,4) to (-1,2)    ?
       SQRT ((-1-(-3))^2 + (2-0)^2) =  SQRT(8)          SQRT ((-1-1)^2 + (2-4)^2) = SQRT(8)
                                                         yes, (-1,2) is equidistant.

  Is the distance from (-3,0) to (-6,7) equal to the distance from (1,4) to (-6,7)   ?
       SQRT ((-6-(-3))^2 + (7-0)^2)           SQRT ((-6-1)^2 + (7-4)^2)      ?
                  SQRT(-3)^2 + 7^2)             SQRT( (-7)^2 + 3^2 )  ?
                       SQRT(58)                               SQRT(58)          yes, (-6,7) is equidistant.

     Is the distance from (-1,2) to (-1,7) equal to 5     ?
            SQRT ( (-1-(-1))^2 + (7-2)^2 )  = 5   ?
                SQRT( 0^2 + 5^2) = 5  ?      yes.  Distance is 5.

     Is the distance from (-6,7) to (-1,7) equal to 5     ?
            SQRT ( (-1-(-6))^2 + (7-7)^2 ) = 5   ?
                  SQRT (5^2) = 5  ?       yes.  Distance is 5

Answer 2

Equidistant from (-3,0) and (1,4) means it will lie on the perpendicular line passing through the mid point of these two points.
mid point = ( (-3+1)/2, (0+4)/2 )
              = (-1 , 2 )

slope of line joining (-3,0) and (1,4) = (y2-y1) / (x2-x1)

                                                     = (4-0 ) /  (1-(-3))

                                                     = 4/4

                                                     = 1

their slope of the perpendicular line = -1/ m  = -1/1 = -1

 

and since this line passes through mid point ( -1,2) so it equation will be  :

 

 

 y -y1 = m (x-x1)                                                                                      

 y - 2 = -1 (x-(-1))

 y - 2 =  -1( x+1)

 y-2 = -x -1

 x+ y = 1 (after rearranging)   eq1.

 

 

now we need to find a point on this line which is at a distance 5 from (-1,7).

let this point be (x',y')

  then    √ { (x'-(-1))²   + (y'-7)² } = 5

                       (x'+1)² + (y'-7)²    = 25    

 solving this eqn. and substituting the values from eq 1. we can find out  x' and y' which is our required solution.

 

            

                    

                                                     

Answer 3

This is a multi part problem.

 

First, we need to find all points equidistant from (-3,0) and (1,4).

 

We'll use the distance formula and set the distances equal implying:

 

sqrt( (x+3)² + (y-0)² ) = sqrt( (x-1)² + (y-4)² ) .........Square both sides

 

(x+3)² + (y-0)² ) = (x-1)² + (y-4)² ......................expand the squared binomials

 

x²+6x+9+y² = x²-2x+1+y²-8y+16 ..............combine like terms, note the x² and y² terms cancel out

 

6x+9 = -2x-8y+17............further simplify and we get the equation of the line

 

x+y=1 ; all points on this line are equidistant fom points (-3,0) and

(1,4)

 

Secondly, we want our points to be at a distance of 5 from (-1,7). This is a circle with radius r = 5 centered at (-1,7).

 

The formula for a circle centered at (h,k) with radius r is: (x-h)²+(y-k)² = r² so for our circle

 

(x+1)² + (y-7)² = 25

 

We need to find where our line x + y =1 intersects the above circle.

Restate the line algebraically as y = 1-x, and substitute this expression into our circle equation for y:

 

(x+1)² + (1-x-7)² = 25 , now we'll expand, collect terms and solve for x:

 

x²+2x+1+x²+12x+36 = 25

 

2x² + 14x +12 = 0

 

2(x² +7x + 6) = 0

 

2(x+6)(x+1) = 0

 

So, x = -6, or x = -1, and we know that y = 1-x, so the points are (-6, 7) and (-1, 2). These are the points that are both on the line x+y = 1 (and equidistant from our two points) and on the circle centered at (-1,7) with a radius of 5.

 

Hope this helps....................................