Given that the equation of the two lines l1 and l2 L1:2x+y =8 L2=6y−mx=3 State the gradiant of l1 and l2 Fine the value of m where l1and l2 are parallel to each other Perpendicular to each other

The equation of $L_1$ is $2x + y = 8$.
To find the gradient, we rearrange it into the slope-intercept form $y = mx + c$, where $m$ is the gradient.

$$y = -2x + 8$$

The gradient of $L_1$ is $m_1 = -2$.

The equation of $L_2$ is $6y - mx = 3$.
Rearrange it into the slope-intercept form $y = mx + c$.

$$6y = mx + 3$$ $$y = \frac{m}{6}x + \frac{3}{6}$$ $$y = \frac{m}{6}x + \frac{1}{2}$$

The gradient of $L_2$ is $m_2 = \frac{m}{6}$.

If two lines are parallel, their gradients are equal ($m_1 = m_2$).

$$-2 = \frac{m}{6}$$

Multiply both sides by 6:

$$m = -2 \times 6$$ $$m = -12$$

The value of $m$ when $L_1$ and $L_2$ are parallel is $\boxed{-12}$.

If two lines are perpendicular, the product of their gradients is -1 ($m_1 \cdot m_2 = -1$).

$$-2 \cdot \frac{m}{6} = -1$$ $$-\frac{2m}{6} = -1$$ $$-\frac{m}{3} = -1$$

Multiply both sides by -3:

$$m = (-1) \times (-3)$$ $$m = 3$$

The value of $m$ when $L_1$ and $L_2$ are perpendicular is $\boxed{3}$.