# Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

**Solution:**

Let (a, 0) be the point on the X- axis that is equidistant from the points (7, 6) and (3, 4).

Using distance formula,

√(7 - a)² + (6 - 0)² = √(3 - a)² + (4 - 0)²

⇒ √49 + a² - 14a + 36 = √9 + a² - 6a + 16

⇒ √a² - 14a + 85 = √a² - 6a + 25

On squaring on both sides, we obtain

a^{2} - 14a + 85 = a^{2} - 6a + 25

⇒ - 14a + 6a = 25 - 85

⇒ - 8a = 60

⇒ a = 60/8

⇒ a = 15/2

Thus, the required point on the x-axis is (15/2, 0)

NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.1 Question 4

## Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4)

**Summary:**

The point on the x-axis, which is equidistant from the points (7, 6) and (3, 4) is (15/2, 0)

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