Question

Coordinates of a point at unit distance from the lines $3x-4y+1=0$ and $8x+6y+1=0$ are

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

Answer 1

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

The correct options are
A $(-\frac{2}{5},-\frac{13}{10})$
B $(\frac{6}{5},-\frac{1}{10})$
C $(0,\frac{3}{2})$
D $(-\frac{8}{5},\frac{3}{10})$

$\frac{A_{1}x+B_{1}y+C_1}{\sqrt{A_1^2+B_1^2}}=\pm \frac{A_{2}x+B_{2}y+C_2}{\sqrt{A_2^2+B_2^2}}$

$\left|\frac{3x-4y+1}{\sqrt{3^2+4^2}}\right|=1\Rightarrow \left|\frac{3x-4y+1}{5}\right|=1$

$\Rightarrow 3x-4y=4$ -----------(1)

Or $3x-4y=-6$ -----------(2)

$\left|\frac{8x+6y+1}{\sqrt{8^2+6^2}}\right|=1\Rightarrow \left|\frac{8x+6y+1}{10}\right|=1$

$\Rightarrow 8x+6y=9$ ----------(3)

Or $8x+6y=-11$ ---------(4)

Solving (1) and (3), we get $x=\frac{6}{5}$ and $y=\frac{-1}{10}$

Solving (2) and (3), we get $x=0$ and $y=\frac{3}{2}$

Solving (1) and (4), we get $x=\frac{-2}{5}$ and $y=\frac{-13}{10}$

Solving (2) and (4), we get $x=\frac{-8}{5}$ and $y=\frac{3}{10}$