Find the coordinates of all points whose distance from (0,-3) is \sqrt{5} and whose distance from (3,-4) is \sqrt{5}?

1 Answer
Feb 15, 2018

the solution is
P-=(1,-5) and P-=(2,-2)

Explanation:

Let P≡(x,y) be the point
Given that P≡(x,y) is away from A≡(0,-3) by sqrt5
and that P≡(x,y) is away from B≡(3,-4) by sqrt5
A and B are separated by the distance given by the relation
(3-0)^2+(-4-(-3))^2=3^2+1^2=10
sqrt10=3.161
while,
sqrt5=2.236
Thus,
sqrt10<2sqrt5

ie
By applying the distance given by the relation
(0-x)^2+(-3-y)^2=5
and
(3-x)^2+(-4-y)^2=5
Simplifying

x^2+(3+y)^2=5
(3-x)^2+(4+y)^2=5

Equating the two distances
x^2+(y+3)^2=(x-3)^2+(y+4)^2
Expanding
x^2+y^2+6y+9=x^2-6x+9+y^2+8y+16
6y-8y+6x+9-9-16=0
Rearanging
6x-2y-16=0
or
3x-y-8=0
The points lie on the straight line
y=3x-8
Substituting in x^2+(y+3)^2=5
x^2+(3x-8+3)^2=5
x^2+(3x-5)^2=5
x^2+9x^2-30x+25=5
10x^2-30x+25-5=0
10x^2-30x+20=0
x^2-3x+2=0

Factorising
x^2-2x-x+2=0
x(x-2)-1(x-2)=0
(x-1)(x-2)=0
x-1=0 or x-2=0
x=1 or x=2
When x=1
y=3x-8becomes
y=3(1)-8=3-8=-5
(x,y)=(1,-5)

When x=2
y=3x-8becomes
y=3(2)-8=6-8=-2
(x,y)=(2,-2)

Check:
For P(1,-5)
Distance from A≡(0,-3)toP≡(1,-5) is given by
sqrt((0-1)^2+(-3-(-5))^2)
=sqrt((-1)^2+2^2)=sqrt(1+4)=sqrt5

For P(2,-2)
Distance from A≡(0,-3)toP≡(2,-2) is given by
sqrt((0-2)^2+(-3-(-2))^2)
=sqrt((-2)^2+(-1)^2)=sqrt(4+1)=sqrt5
Hence,

the solution is
P-=(1,-5) and P-=(2,-2)