Prev

Next

Q39 of 51 Page 10

Find the points on the x-axis, whose distances from the line = 1 are 4 units.

Let the point be (x, 0)

Now = 1

or 4x + 3y = 12

or 4x + 3y – 12 = 0

Therefore d =

4 =

or 10 =

Now 4x₁ = 12 or 4x₁ = 12 = -20

4x₁ = 32 or 4x₁ = 8

x₁ = 8 or x₁ = -2

Hence required points are (8, 0) and (-2, 0)

More from this chapter

37

Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x- axis.(i) x -

y+ 8 = 0 (ii) y – 2 = 0 (iii) x – y = 4

38

Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y – 2).

40

Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

(ii) l(x + y) + p = 0 and lx + ly – r = 0

41

Fine the equation of the line parellel to the line 3x-4y+2=0 and passing through the point (-2,3)