If the distance between the points (k, 0) and (0, k) is 5 units, then find the value of k.

Using the distance formula with points A(k, 0) and B(0, k), we have: square root of (0 - k)² + (k - 0)² equals 5. Simplifying: square root of k² + k² equals 5, which gives square root of 2k² equals 5. Therefore k times square root of 2 equals 5, so k equals 5 divided by square root of 2. Rationalizing: k equals 5 times square root of 2 divided by 2. This can also be written as k equals plus or minus 5 divided by square root of 2, giving two values.

Class 10 Maths Chapter 7 Exercise 7.1 Question 9 – If Distance Between Points is 5, Find k

🖼️ Visual Representation

Figure: Points A(k,0) and B(0,k) are at distance 5 units from each other

📋 Given

Point A: \((k, 0)\) — lies on the x-axis
Point B: \((0, k)\) — lies on the y-axis
Distance AB: 5 units

💡 Notice: Both points have the same value ‘k’ but in different positions (one on x-axis, one on y-axis).

🎯 To Find

The value of k for which the distance between A and B is exactly 5 units.

📐 Formula

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

📝 Solution

1 Apply the distance formula:

Using points \(A(k, 0)\) and \(B(0, k)\):

\[ AB = \sqrt{(0 – k)^2 + (k – 0)^2} \] \[ 5 = \sqrt{(-k)^2 + k^2} \]

2 Simplify the expression:

Since \((-k)^2 = k^2\):

\[ 5 = \sqrt{k^2 + k^2} \] \[ 5 = \sqrt{2k^2} \]

3 Simplify the square root: \[ 5 = \sqrt{2} \cdot \sqrt{k^2} \] \[ 5 = \sqrt{2} \cdot |k| \]

⚠️ Remember: \(\sqrt{k^2} = |k|\) (absolute value of k)

4 Solve for |k|: \[ |k| = \frac{5}{\sqrt{2}} \]

5 Rationalize the denominator:

Multiply numerator and denominator by \(\sqrt{2}\):

\[ |k| = \frac{5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \] \[ |k| = \frac{5\sqrt{2}}{2} \]

6 Find both values of k:

Since \(|k| = \frac{5\sqrt{2}}{2}\), we have two solutions:

\[ k = +\frac{5\sqrt{2}}{2} \quad \text{or} \quad k = -\frac{5\sqrt{2}}{2} \]

💡 Both positive and negative values are valid because distance is always positive.

✅ Answer: \( k = \pm\frac{5\sqrt{2}}{2} \)

or approximately: k ≈ ±3.54 units

✓ Verification

Let’s verify both values:

Value of k	Point A	Point B	Distance
\(k = \frac{5\sqrt{2}}{2}\)	\((\frac{5\sqrt{2}}{2}, 0)\)	\((0, \frac{5\sqrt{2}}{2})\)	\(\sqrt{2 \times (\frac{5\sqrt{2}}{2})^2} = 5\) ✓
\(k = -\frac{5\sqrt{2}}{2}\)	\((-\frac{5\sqrt{2}}{2}, 0)\)	\((0, -\frac{5\sqrt{2}}{2})\)	\(\sqrt{2 \times (\frac{5\sqrt{2}}{2})^2} = 5\) ✓

Both values satisfy the condition!

💡 Geometric Interpretation

Understanding the Problem Visually:

Point A lies on the x-axis at distance |k| from origin
Point B lies on the y-axis at distance |k| from origin
The line segment AB forms the hypotenuse of a right triangle with legs of length |k| each
By Pythagoras theorem: \(AB^2 = k^2 + k^2 = 2k^2\)
Given AB = 5, so \(25 = 2k^2\), which gives \(k^2 = \frac{25}{2}\)

Two Cases:

Case 1 (k positive): Points in 1st quadrant region
Case 2 (k negative): Points in 3rd quadrant region

In both cases, the distance between the points remains 5 units!

⚠️ Common Mistakes

❌ Mistake 1: Forgetting that \(\sqrt{k^2} = |k|\), not just k.
✅ Correct: Always consider both positive and negative values.

❌ Mistake 2: Not rationalizing the denominator in the final answer.
✅ Correct: \(\frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}\) is the standard form.

❌ Mistake 3: Calculation error: \((-k)^2 = -k^2\).
✅ Correct: \((-k)^2 = k^2\) (always positive).

❌ Mistake 4: Thinking only positive k is valid.
✅ Correct: Both positive and negative k give the same distance.

💡 Key Points

Points on x-axis have form \((x, 0)\)
Points on y-axis have form \((0, y)\)
\(\sqrt{k^2} = |k|\) (absolute value)
When \(|k| = a\), then \(k = \pm a\)
Always rationalize denominators containing square roots
Distance is always positive, but coordinates can be negative
This problem demonstrates symmetry in coordinate geometry
\(\sqrt{2k^2} = k\sqrt{2}\) when k is positive, or \(|k|\sqrt{2}\) in general