The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis at
a. (0, 13)
b. (0, –13)
c. (0, 12)
d. (13, 0)

Solution:

The points given are

A(x₁, y₁) = (1, 5)

B(x₂, y₂) = (4, 6)

Perpendicular bisector of AB will pass through the mid-point of AB

Perpendicular bisector of AB will meet the Y axis at P(0, y)

AP = BP

We know that

The distance between two points P(x₁, y₁) and Q(x₂, y₂) is

\(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

Let us apply the distance formula

AP² = BP²

(x₁ - 0)² + (y₁ - y)² = (x₂ - 0)² + (x₂ - y)²

Substituting the values

(1- 0)² + (5 - y)² = (4 - 0)² + (6 - y)²

Let us expand using (a - b)² = a² + b² - 2ab

1 + 25 + y² - 10y = 16 + 36 + y² - 12y

By further calculation

26 + y² - 10y = 52 + y² - 12y

So we get

y² - 10y - y² + 12y + 26 - 52 = 0

2y - 26 = 0

2y = 26

Divide both sides by 2

y = 13

Therefore, the perpendicular bisector of the line segment is (0, 13).

✦ Try This: The perpendicular bisector of the line segment joining the points A (2, 6) and B (5, 6) cuts the y-axis at

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

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NCERT Exemplar Class 10 Maths Exercise 7.1 Problem 14

The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis at a. (0, 13), b. (0, –13), c. (0, 12), d. (13, 0)

Summary:

The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis at (0, 13)

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