analytic geometry?

Question

Find the point equidistant from (-6,-1) and (-1,2) and at a distance 5 from (-2,7)

Answer 1

Hi,

To find all the points that are equidistant from (-6,-1) and (-1,2), we must find the equation of the perpendicular bisector of the segment connecting these 2 points. It goes through the midpoint of the segment. This midpoint is found from xm = (x1 + x2)/2 and ym = (y1 + y2)/2. Our midpoint is found by xm = (-6 + -1)/2 = -3.5 and ym = (-1 + 2)/2 = .5. The midpoint is at
(-3.5,.5).

Next we need to know the slope of the segment connecting
(-6,-1) and (-1,2). It is found by m = (y2 - y1)/(x2 - x1). The slope is m = (2 - (-1))/(-1 - (-6)) = 3/5. The slope of any perpendicular is its negative reciprocal, -5/3.

Using the slope of -5/3 for a perpendicular and the point (-3.5,.5), the perpendicular bisector line is given by the point-slope equation y - ½ = -5/3(x + 7/2). This simplifies to
y = -5/3x - 16/3.

If we are looking for a point that is 5 units from (-2,7), that point is on a circle with a center of (-2,7) with a radius of 5.
Its equation is (x + 2)² + (y - 7)² = 25.

The point that is both equidistant from both (-6,-1) and (-1,2) and also 5 units from (-2,7) must be on both the perpendicular bisector equation, y = -5/3x - 16/3, and on the circle's equation, (x + 2)² + (y - 7)² = 25.

If we expand the circle's equation, it becomes:

x² + 4x + 4 + y² - 14y + 49 = 25

This simplifies to:

x² + 4x + y² - 14y + 28 = 0

If we substitute -5/3x - 16/3 from the other equation in place of y. the equation becomes:

x² + 4x + (-5/3x - 16/3)² - 14(-5/3x - 16/3) + 28 = 0

x² + 4x + (-5/3x - 16/3)(-5/3x - 16/3) - 14(-5/3x - 16/3) + 28 = 0

x² + 4x + 25x²/9 + 160x/9 + 256/9 +70x/3 + 224/3 + 28 = 0

Multiplying by 9 to eliminate fractions and then combining terms gives the equation:

9x² + 36x + 25x² + 160x + 256 + 210x + 672 + 252 = 0
34x² + 406x + 1180 = 0

Factoring gives:
2(17x² + 203x + 590) = 0
2(17x + 118)(17x + 85) = 0

Solving one factor for x gives:
17x + 85 = 0
17x = -85
x = -5

If we substitute -5 for x and solve for y, we get:
y = -5/3x - 16/3
y = -5/3(-5) - 16/3
y = 25/3 - 16/3
y = 9/3 or 3

One point equidistant from (-6,-1) and (-1,2) and 5 units away from (-2,7) is (-5,3).

Solving the other factor for x:
17x + 118 = 0
17x = -118/17

If we substitute -118/17 for x and solve for y, we get:
y = -5/3x - 16/3
y = -5/3(-118/17) - 16/3
y = 590/51 - 16/3
y = 106/17

So the point (-118/17,106/17) is also equidistant from (-6,-1) and (-1,2) and 5 units away from (-2,7) is (-5,3).

So there are 2 answers: (-5,3) and (-118/17,106/17).

I hope that helps!! :-)

Answer 2

The locus of points equidistant from the two given points is a line thru and perpendicular to the midpoint of the two given points.

Calculate the midpoint.

M = [(-6-1)/2, (-1+2)/2] = M(-7/2, 1/2)

The slope is the negative reciprocal of the slope of the line thru the two given points.

m = (2 + 1)/(-1 + 6) = 3/5
m' = -1/m = -5/3

The desired point lies on the line:

y - 1/2 = (-5/3)(x + 7/2)
y = (-5/3)x - 35/6 + 1/2
y = (-5/3)x - 16/3

The locus of points a distance of 5 from the P(-2, 7) is the circle:

(x + 2)² + (y - 7)² = 25

Find the intersection of the circle and the line.

(x + 2)² + (y - 7)² = 25
(x + 2)² + [(-5/3)x - 16/3 - 7]² = 25
(x + 2)² + [(-5/3)x - 37/3]² = 25
x² + 4x + 4 + (25/9)x² + (370/9)x + 1369/9 - 25 = 0
(34/9)x² + (406/9)x + 1180/9 = 0

34x² + 406x + 1180 = 0
17x² + 203x + 590 = 0
(17x + 118)(x + 5) = 0
x = -118/17, -5

y = (-5/3)x - 16/3
y = 318/51, 3

The two points are:

P(-118/17, 318/51) and Q(-5,3)

Answer 3

The line between (-6,-1)and (-1,2) has a slope of 3/5. Thus its perpendicular bisector has a slope of -5/3. The midpoint of the line is (-3.5, 1/2). So it's equation is y= -5/3x + b
1/2 = (-5/3)( -7/2)+b
1/2- 35/6 = b
b = -32/6 =-16/3
So equation of perpendicular bisector is y = -5/3x -16/3

Now we must see if the point (-2,7) is a distance of 5 from the perpendicular bisector.

The line from (-2,7) to the perpendicular bisector has the equation y = 3/5x + b
7 = 3/5 (-2) + b so b = 8 1/5
So its equation is y = 3/5x + 8 1/5 These lines intersect at thepoint (-5.97,4.61) .
The distance between this point and (-2,7) is given by:
sqrt((-2-(-5.97))^2 + (7- (4.61))^2) = 4.63

So it looks like there is no such point, unless I have made an arithmetic error somewhere.