Straight Line And Pair Of Straight Lines Ques 24
- 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?
(2019 Main, 10 April I)
(a) $-\frac{1}{4},-\frac{2}{3}$
(b) $-\frac{1}{4}, \frac{2}{3}$
(c) $\frac{1}{4},-\frac{1}{3}$
(d) $\frac{1}{4}, \frac{1}{3}$
Show Answer
Solution:
- Since, equation of a line parallel to line $a x+b y+c=0$ is $a x+b y+k=0$
$\therefore$ Equation of line parallel to line
$$ 4 x-3 y+2=0 \text { is } 4 x-3 y+k=0 $$
Now, distance of line (i) from the origin is
$$ \frac{|k|}{\sqrt{4^{2}+3^{2}}}=\frac{3}{5} $$
$$ \begin{aligned} \Rightarrow & & |k| & =3 \\ \Rightarrow & & k & = \pm 3 \end{aligned} $$
So, possible lines having equation, either $4 x-3 y+3=0$ or $4 x-3 y-3=0$
Now, from the given options the point $-\frac{1}{4}, \frac{2}{3}$ lies on the line $4 x-3 y+3=0$.
BETA
