Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units. 106
Step-by-Step Solution
Key Concept: Use the distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\): \[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.\] Set the distance equal to 10 and solve the resulting quadratic equation for \(y\).
1. Write down the coordinates of the two points:
\[ P(2, -3) \quad \text{and} \quad Q(10, y). \]
2. Apply the distance formula:
\[ \sqrt{(10-2)^2 + (y-(-3))^2} = 10. \]
3. Simplify inside the square root:
\[ \sqrt{8^2 + (y+3)^2} = 10 \]
\[ \sqrt{64 + (y+3)^2} = 10. \]
4. Square both sides to remove the square root (both sides are non‑negative):
\[ 64 + (y+3)^2 = 100. \]
5. Isolate the quadratic term:
\[ (y+3)^2 = 100 - 64 = 36. \]
6. Take square roots (remember both positive and negative roots):
\[ y+3 = \pm 6. \]
7. Solve for \(y\):
- If \(y+3 = 6\), then \(y = 6 - 3 = 3\).
- If \(y+3 = -6\), then \(y = -6 - 3 = -9\).
8. Hence the required values of \(y\) are \(y = 3\) and \(y = -9\).
Correct Answer: y = 3 \;\text{or}\; y = -9