If \(A(2,-4)\) is equidistant from \(P(3,8)\) and \(Q(-10,y)\), find \(y\). Also find \(PQ\).
\(y=-3\) or \(y=-5\)
\(PQ=\sqrt{290}\) when \(y=-3\); and \(PQ=13\sqrt{2}\) when \(y=-5\).
Step 1: Since point \(A(2,-4)\) is equidistant from \(P(3,8)\) and \(Q(-10,y)\), it means:
\(AP = AQ\)
Step 2: Use the distance formula:
\(AP = \sqrt{(2-3)^2 + (-4-8)^2}\)
\(AQ = \sqrt{(2-(-10))^2 + (-4-y)^2}\)
Step 3: Square both sides to remove the square root:
\((2-3)^2 + (-4-8)^2 = (2+10)^2 + (-4-y)^2\)
Step 4: Simplify each term:
So the equation becomes:
\(1 + 144 = 144 + (y+4)^2\)
\(145 = 144 + (y+4)^2\)
Step 5: Subtract 144 from both sides:
\((y+4)^2 = 1\)
Step 6: Take square root:
\(y+4 = ±1\)
So, \(y = -4+1 = -3\) or \(y = -4-1 = -5\)
Step 7: Now find \(PQ\) using the distance formula.
For \(y = -3\):
\(PQ = \sqrt{(3 - (-10))^2 + (8 - (-3))^2} = \sqrt{13^2 + 11^2} = \sqrt{169 + 121} = \sqrt{290}\)
For \(y = -5\):
\(PQ = \sqrt{(3 - (-10))^2 + (8 - (-5))^2} = \sqrt{13^2 + 13^2} = \sqrt{169 + 169} = \sqrt{338}\)
Final Answer: \(y = -3\) or \(y = -5\). \(PQ = \sqrt{290}\) when \(y = -3\), and \(PQ = \sqrt{338}\) when \(y = -5\).