What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units

Solution:

Let (0, b) be the point on y-axis whose distance from line x/3 + y/4 = 1 is 4 units.

The given line can be written as

4x + 3y - 12 = 0 ....(1)

On comparing equation (1) to the general equation of line Ax + By + C = 0 , we obtain

A = 4, B = 3 and C = - 12.

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x\(_1\), y\(_1\)) is given by

d = |Ax\(_1\) + By\(_1\) + C|/√A² + B²

Therefore, if (0, b) is the point on the y-axis whose distance from line x/3 + y/4 = 1 is 4 units,

Then,

⇒ 4 = |4(0) + 3(b) - 12|/√4² + 3²

⇒ 4 = |3b - 12|/5

⇒ 20 = |3b - 12|

⇒ 20 = ± (3b - 12)

⇒ (3b - 12) = 20 or (3b - 12) = - 20

⇒ 3b = 20 + 12 or 3b = - 20 + 12

⇒ b = 32/3 or b = - 8/3

Thus, the required points are (0, 32/3) and (0, - 8/3)

NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 4

Summary:

The points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units are (0, 32/3) and (0, - 8/3)

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