NCERT Solutions
Class 11 Maths
Straight Lines

Ex.Misc. Q.4
What are the points on the y-axis whose distance from the line +
= 1 is 4 units.
Let (0, b) be the point on the y-axis whose distance from line +
= 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 get
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 (x1, y1) is
given by d = |Ax1 + By1 + C| ÷ √ (A2 + B2)
Therefore, if (0, b) is the point on the y-axis whose distance from line +
= 1 is 4 units, then:
4 = |4 × 0 + 3 × b - 12| ÷ √ (42 + 32)
=> 4 = |3b - 12| ÷ 5
=> 20 = ± (3b – 120)
=> 20 = 3b – 12 or 20 = -(3b – 12)
=> 3b = 32 or 3b = -20 + 12
=> b = or b = -
Thus, the required points are (0, ) or (0, -
).