Lines are drawn parallel to the line 4x - 3y + 2 = 0, at a distance (3/5) from the origin. Then which one of the following points lies on any of these lines ?

Question

Lines are drawn parallel to the line 4x - 3y + 2 = 0, at a distance (3/5) from the origin. Then which one of the following points lies on any of these lines ? (A) ( - (1/4), (2/3))  (B) ( (1/4), (1/3))  (C) ( - (1/4), - (2/3))  (D) ((1/4), - (1/3))

Tardigrade Admin · Accepted Answer

Required line is 4x - 3y + λ = 0 | (λ/5) | = (3/5) ⇒ λ = ± 3 . So, required equation of line is 4x - 3y + 3 = 0 and 4x - 3y - 3 = 0 (1) 4( - (1/4)) - 3 ((2/3)) + 3 = 0

Q. Lines are drawn parallel to the line $4 x - 3 y + 2 = 0$ , at a distance $\frac{3}{5}$ from the origin. Then which one of the following points lies on any of these lines ?

$(- \frac{1}{4}, \frac{2}{3})$

$(\frac{1}{4}, \frac{1}{3})$

$(- \frac{1}{4}, - \frac{2}{3})$

$(\frac{1}{4}, - \frac{1}{3})$

Required line is $4 x - 3 y + λ = 0$
$∣ ∣ \frac{λ}{5} ∣ ∣ = \frac{3}{5}$
$\Rightarrow λ = \pm 3.$
So, required equation of line is $4 x - 3 y + 3 = 0$ and $4 x - 3 y - 3 = 0$
$(1) 4 (- \frac{1}{4}) - 3 (\frac{2}{3}) + 3 = 0$

Q. Lines are drawn parallel to the line 4x−3y+2=0, at a distance 53​ from the origin. Then which one of the following points lies on any of these lines ?

(−41​,32​)

(41​,31​)

(−41​,−32​)

(41​,−31​)