Question

Statement I The equation of the line which is equidistant from parallel line $9x+6y-7=0$ and $3x+2y+6=0$ is $18x+12y+11=0$.
Statement II Any line parallel to $ax+by+c=0$ is $ax+by+k=0$ and distance between two lines is $\left|\frac{c-k}{\sqrt{a^2+b^2}}\right|$

• A. Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I
• B. Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I
• C. Statement I is true; Statement II is false
• D. Statement I is false; Statement II is true

Answer 1

Answer: (A)

Equations of given lines are
$9x+6y-7=\mathrm{0.....}$(i)
or $3x+2y-\frac{7}{3}=0$
and $3x+2y+6=\mathrm{0.....}$(ii)
Equation of line parallel to either Eq. (i) or Eq. (ii), is
$3x+2y+k=\mathrm{0....}$(iii)
Now, given line (iii) is equidistant from lines (i) and (ii).

$\therefore$ Distance between lines (i) and (iii)
$\Rightarrow \left|\frac{k+\frac{7}{3}}{\sqrt{3^2+2^2}}\right|=\left|\frac{k-6}{\sqrt{3^2+2^2}}\right|$
$\Rightarrow (\because$ distance between parallel lines is $\left|\frac{c-d}{\sqrt{a^2+b^2}}\right|)$
$\Rightarrow \left|k+\frac{7}{3}\right|=|k-6|$
$k+\frac{7}{3}=\pm (k-6)$
Taking negative sign,
$\Rightarrow k+\frac{7}{3}=-k+6\Rightarrow 2k=6-\frac{7}{3}$
$\Rightarrow 2k=\frac{11}{3}\Rightarrow k=\frac{11}{6}$
Hence, Eq. (iii) becomes
$3x+2y+\frac{11}{6}=0\left(\text{ put }k=\frac{11}{6}\right)$
$\therefore 18x+12y+11=0$ is the required line.