If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Step-by-Step Solution
Key Concept: Use the distance formula \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). Since Q is equidistant from P and R, set \(QP = QR\) and solve for \(x\). Then compute the required distances using the same formula.
1. Write the distance QP using the distance formula:
$$QP = \sqrt{(5-0)^2+(-3-1)^2}=\sqrt{5^2+(-4)^2}=\sqrt{25+16}=\sqrt{41}.$$
2. Express QR in terms of the unknown \(x\):
$$QR = \sqrt{(x-0)^2+(6-1)^2}=\sqrt{x^2+5^2}=\sqrt{x^2+25}.$$
3. Equate the two distances because Q is equidistant from P and R:
$$\sqrt{x^2+25}=\sqrt{41}\;\Rightarrow\;x^2+25=41\;\Rightarrow\;x^2=16.$$
4. Solve for \(x\):
$$x = \pm 4.$$
5. Find QR (which equals QP):
$$QR = \sqrt{41} \;\text{units}.$$
6. Find PR for each value of \(x\):
- For \(x = 4\):
$$PR = \sqrt{(4-5)^2+(6-(-3))^2}=\sqrt{(-1)^2+9^2}=\sqrt{1+81}=\sqrt{82}.$$
- For \(x = -4\):
$$PR = \sqrt{(-4-5)^2+(6-(-3))^2}=\sqrt{(-9)^2+9^2}=\sqrt{81+81}=\sqrt{162}=9\sqrt{2}.$$
Correct Answer: x = 4 or x = -4; QR = \sqrt{41} units; PR = \sqrt{82} units when x = 4, and PR = 9\sqrt{2} units when x = -4.