Find a point equidistant from \(A(-5,4)\) and \(B(-1,6)\). How many such points are there?
One example is (-3, 5). There are infinitely many such points (on the perpendicular bisector).
Step 1: Let \(P(x,y)\) be any point equidistant from \(A(-5,4)\) and \(B(-1,6)\).
Then, \(PA = PB\). To avoid square roots, we use squares:
\(PA^2 = PB^2\).
Step 2: Write \(PA^2\) and \(PB^2\) using distance formula.
\(PA^2 = (x-(-5))^2 + (y-4)^2 = (x+5)^2 + (y-4)^2\)
\(PB^2 = (x-(-1))^2 + (y-6)^2 = (x+1)^2 + (y-6)^2\)
Step 3: Equate and expand.
\((x+5)^2 + (y-4)^2 = (x+1)^2 + (y-6)^2\)
Expand each term:
So,
\(x^2 + 10x + 25 + y^2 - 8y + 16 = x^2 + 2x + 1 + y^2 - 12y + 36\)
Step 4: Cancel common terms and simplify.
Cancel \(x^2\) and \(y^2\) from both sides:
\(10x + 41 - 8y = 2x + 37 - 12y\)
Bring all terms to one side:
\(10x - 2x + 41 - 37 + (-8y + 12y) = 0\)
\(8x + 4 + 4y = 0\)
Divide by 4:
\(2x + y + 1 = 0\)
Step 5: Interpret the result.
The equation \(2x + y + 1 = 0\) represents a straight line. Every point on this line is equidistant from A and B (this line is the perpendicular bisector of \(AB\)).
Hence, there are infinitely many such points.
Step 6: Give one such point (example).
One easy point is the midpoint of \(AB\):
\(M\left(\dfrac{-5+(-1)}{2},\dfrac{4+6}{2}\right) = \left(\dfrac{-6}{2},\dfrac{10}{2}\right) = (-3,5)\)
So, \((-3,5)\) is equidistant from A and B.