Find the distance of the line 4x + 7 y + 5 = 0 from the point (1, 2) along the line 2x - y = 0

Solution:

The given lines are

2x - y = 0 ....(1)

4x + 7y + 5 = 0 ....(2)

Let A(1, 2) is a point on the line (1) and B be the point intersection of lines (1) and (2).

On solving equations (1) and (2), we obtain x = - 5/18 and y = - 5/9

Coordinates of point B are (- 5/18, - 5/9)

By using distance formula, the distance between points A and B can be obtained as

AB = √(1 + 5/18)² + (2 + 5/9)²

= √(23/18)² + (23/9)²

= √(23/9 x 2)² + (23/9)²

= √(23/9)² (1/2)² + (23/9)²

= √(23/9)² (1/4 + 1)

= 23/9 √5/4

= 23/9 x √5/2

= 23√5/18

Thus, the required distance is 23√5/18 units

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NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 15

Find the distance of the line 4x + 7 y + 5 = 0 from the point (1, 2) along the line 2x - y = 0

Summary:

The distance of the line 4x + 7 y + 5 = 0 from the point (1, 2) along the line 2x - y = 0 is 23√5/18 units