Question

The coordinates of a point at unit distance from the lines 3x - 4y + 1 = 0 and 3x + 6y + 1 = 0 are

• A. $(\frac{6}{5},\frac{-1}{10})$
• B. $(0,\frac{-1}{10})$
• C. $(\frac{-2}{5},\frac{-13}{10})$
• D. $(\frac{-8}{5},\frac{3}{10})$

Answer 1

Answer: (A), (C), (D)

The correct options are
A

$(\frac{6}{5},\frac{-1}{10})$

C

$(\frac{-2}{5},\frac{-13}{10})$

D

$(\frac{-8}{5},\frac{3}{10})$

Let (x, y) be the coordinate of the point.
$\frac{3x-4y+1}{\sqrt{3^2+4^2}}=\pm 1and\frac{8x+6y+1}{\sqrt{8^2+6^2}}=\pm 1$
3x - 4y - 4 = 0 . . . (1)
3x - 4y + 6 = 0 . . . (2)
8x + 6y - 9 = 0 . . . (3)
8x + 6y + 11 = 0 . . . (4)
Solving (1) and (3) we get (x,y) $=(\frac{6}{5},\frac{-1}{10})$.
(1) and (4) we get (x, y) $=(\frac{-2}{5},\frac{-13}{10})$
(2) and (3) we get (x, y) $=(0,\frac{3}{2})$
and (2) and (4) we get (x, y) $=(\frac{-8}{5},\frac{3}{10})$