Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.

Using the distance formula with P(2, -3) and Q(10, y), we have: square root of (10 - 2)² + (y - (-3))² equals 10. Simplifying: square root of 64 + (y + 3)² equals 10. Squaring both sides: 64 + (y + 3)² equals 100. Therefore (y + 3)² equals 36. Taking square root: y + 3 equals plus or minus 6. This gives two values: y equals 3 or y equals -9.

Class 10 Maths Chapter 7 Exercise 7.1 Question 8 – Find Values of y for Given Distance

📋 Given

Point P: \((2, -3)\)
Point Q: \((10, y)\)
Distance PQ: 10 units

🎯 To Find

The values of y for which the distance between P and Q is 10 units.

📐 Formula

\[ \text{Distance} = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \]

The distance formula calculates the straight-line distance between two points in a coordinate plane.

📝 Solution

1 Apply the distance formula:

Using points \(P(2, -3)\) and \(Q(10, y)\):

\[ PQ = \sqrt{(10 – 2)^2 + (y – (-3))^2} \] \[ 10 = \sqrt{8^2 + (y + 3)^2} \] \[ 10 = \sqrt{64 + (y + 3)^2} \]

2 Square both sides to eliminate the square root: \[ 10^2 = 64 + (y + 3)^2 \] \[ 100 = 64 + (y + 3)^2 \]

3 Isolate the squared term: \[ (y + 3)^2 = 100 – 64 \] \[ (y + 3)^2 = 36 \]

4 Take square root of both sides: \[ y + 3 = \pm\sqrt{36} \] \[ y + 3 = \pm 6 \]

⚠️ Remember: When taking the square root, we get both positive and negative values.

5 Solve for y (two cases):

Case 1: When \(y + 3 = 6\)

\[ y = 6 – 3 = 3 \]

Case 2: When \(y + 3 = -6\)

\[ y = -6 – 3 = -9 \]

✅ Answer: y = 3 or y = -9

✓ Verification

Let’s verify both values by calculating the actual distances:

Value of y	Point Q	Distance Calculation	Result
y = 3	(10, 3)	\(\sqrt{64 + 36} = \sqrt{100}\)	10 units ✓
y = -9	(10, -9)	\(\sqrt{64 + 36} = \sqrt{100}\)	10 units ✓

Both values are correct! This means there are two possible positions for point Q on the coordinate plane, both at a distance of 10 units from point P.

💡 Geometric Interpretation

Geometrically, this problem asks: “What are the y-coordinates of points on the vertical line x = 10 that are exactly 10 units away from point P(2, -3)?”

The answer gives us two points: Q₁(10, 3) and Q₂(10, -9). Both lie on a circle centered at P with radius 10 units.

Since both points have the same x-coordinate (10), they lie on a vertical line. The two y-values represent the upper and lower intersection points of this vertical line with the circle.

⚠️ Common Mistakes

❌ Mistake 1: Forgetting the ± sign when taking square root.
✅ Correct: Always consider both positive and negative roots.

❌ Mistake 2: Sign error: \(y – (-3) = y – 3\) instead of \(y + 3\).
✅ Correct: Subtracting a negative gives: \(y – (-3) = y + 3\).

❌ Mistake 3: Not verifying both answers.
✅ Correct: Always substitute back to check your solutions.

💡 Key Points

The distance formula is derived from the Pythagoras theorem
When \(a^2 = b^2\), then \(a = \pm b\) (two solutions)
Squaring both sides eliminates square roots but may introduce extraneous solutions (always verify)
Problems involving “find the value” often have multiple solutions
This type of problem is related to finding points on a circle
Always check your answer by substituting back into the original equation