Parallelism !!!!!

It is given that:

$\begin{cases} 3x+4y = 9 & \Rightarrow y = - \frac{3}{4} x + \frac{9}{4} &...(1) \\ 6x + 8y = 15 & \Rightarrow y = - \frac{3}{4} x + \frac{15}{8} &...(2) \end{cases}$

Therefore, the y-axis intercepts of Line 1 and Line 2 are $A(0,\frac{9}{4})$ and $B(0,\frac{15}{8})$ respectively. Let the perpendicular from $B$ to Line 1 to meet at $C$ , then we note that $\angle ABC = \theta$ is same as the angle the two parallel lines make with x-axis. Therefore, $\tan{\theta} = \frac{3}{4}$ and $\triangle ABC$ is a Pythagorean $3$ - $4$ - $5$ triangle.

Therefore, the distance between the two lines:

$BC = \frac{4}{5} \times AC = \frac {4}{5}(\frac{9}{4} - \frac{15}{8} ) = \frac{4}{5}\times \frac{3}{8} = \boxed{\frac{3}{10}}$

The tangent for the first line = -3/4 . and the second line tangent = -3/4 . so the two lines are parallel . the distance between them = (ax+by+c)/ (squrt(a^2+b^2)). By assume the point (1/3 , 2) lies on the line 3x+4y=9 . then by using this point and the second line equation in the distance equation we get that . d=(6×(1÷3)+8×2-15)/(squrt(6^2+8^2))= 3/10