Find the points on the x-axis whose distances from the line x/3 + y/4 = 1 are 4 units

Solution:

The given equation of line is

x/3 + y/4 = 1

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

On comparing equation (1) with general equation of a line Ax + By + C = 0 we obtain A = 4, B = 3 , and C = - 12.

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.

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,

|4(a) + 3(0) -12|/√4² + 3² = 4

⇒ |4a - 12|/5 = 4

⇒ |4a - 12| = 20

⇒ (4a - 12) = 20 or - (4a - 12) = 20

⇒ 4a = 20 + 12 or 4a = - 20 + 12

⇒ a = 8 or a = - 2

Thus, the required points on x-axis are (-2, 0) and (8, 0)

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NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.3 Question 5

Find the points on the x-axis whose distances from the line x/3 + y/4 = 1 are 4 units

Summary:

The points on the x-axis whose distances from the line x/3 + y/4 = 1 are 4 units are (-2, 0) and (8, 0)