Question

The distance between the two lines represented by the equation $9x^2-24xy+16y^2-12x+16y-12=0$ is __ units

• A. $\frac{8}{5}$
• B. $\frac{6}{5}$
• C. $\frac{11}{5}$
• D. $\frac{9}{5}$

Answer 1

The correct option is A $\frac{8}{5}$

$9x^2-24xy+16y^2-12x+16y-12=0\Rightarrow (3x-4y{)}^2-4(3x-4y)-12=0$
$\Rightarrow (3x-4y+2)(3x-4y-6)=0$
So, the lines will be $3x-4y+2=0;3x-4y-6=0$
Hence, distance between lines is
$\frac{|-6-2|}{5}=\frac{8}{5}\text{units}$

Alternate solution:
$\because h^2-ab=144-16\cdot 9=0$
$\therefore$ Given line represents parallel lines
So, distance $=2\sqrt{\frac{g^2-ac}{a(a+b)}}$
$=2\sqrt{\frac{36-9(-12)}{9(9+16)}}=\frac{2\times 12}{15}=\frac{8}{5}$ units