Find the equations of the straight lines passing through the point (−2, −7) and having an intercept of length 3 between

Solution

The equation of any line through the point (−2, −7) is

, …(i)

where | r | is the distance of any point (x, y) on the line (i) from the point (−2, −7).

The given parallel lines are

4x + 3y = 3 …(ii)

and 4x + 3y = 12 …(iii)

Since the line (i) makes an intercept of length 3 between the parallel lines (ii) and (iii), therefore corresponding points on the line (i) will be on the lines (ii) and (iii).

Thus the point (r cosθ − 2, r sinθ − 7) will be on 4x + 3y = 3, and the point

[(r + 3) cosθ − 2, (r + 3) sinθ − 7]

will lie on 4x + 3y = 12.

∴ 4 (r cosθ − 2) + 3 (r sinθ − 7) = 3 …(iv)

and 4[(r + 3) cosθ − 2] + 3[(r + 3) sinθ − 7] = 12 …(v)

Subtracting (iv) from (v), we get

12 cosθ + 9 sinθ = 9 …(vi)

or 12 cosθ = 9 (1 − sinθ)    or 144 cos2θ = 81 (1 − 2sinθ + sin2θ)

or 25sin2θ − 18sinθ − 7 = 0    or (25sinθ + 7) (sinθ − 1) = 0

∴ sinθ = 1, −7/25.

When sinθ = 1, cosθ = 0 and when sinθ = −7/25, cosθ = 24/25, from (vi).

Hence from (i) the required lines are

 and  

i.e., x + 2 = 0  and 7x + 24y + 182 = 0.