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

Smart Achievers · Accepted Answer

(3)

The given line is $4 x - 3 y + 2 = 0$ .

The straight line parallel to the given line is $4 x - 3 y + c = 0$

Distance of this line from origin is $\frac{3}{5}$ .

$∴$ $| \frac{c}{\sqrt{16 + 9}} |$ $= \frac{3}{5} \Rightarrow$ $| \frac{c}{5} |$ $= \frac{3}{5} \Rightarrow c = \pm 3$

So, the parallel lines are $4 x - 3 y + 3 = 0$ or $4 x - 3 y - 3 = 0$ .

Now, point $(- \frac{1}{4}, \frac{2}{3})$ given in option (3) satisfies the equation, $4 x - 3 y + 3 = 0 .$

So, $(- \frac{1}{4}, \frac{2}{3})$ lies on the line $4 x - 3 y + 3 = 0$ .

Solution

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

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

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

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

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

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Solution

Q. Lines are drawn parallel to the line 4x-3y+2=0, at a distance 35 from the origin. Then which one of the following points lies on any of these lines

1 (14,-13)

2 (-14,-23)

3 (-14,23)