Find the distance between the points A(-3, 2) and B(4, -6). Find the acute angle between the lines 2x + 3y = 5 and 3x + 2y = 1 correct to one decimal place.

The distance $D$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Given $A(-3, 2)$ and $B(4, -6)$, we have $x_1 = -3$, $y_1 = 2$, $x_2 = 4$, $y_2 = -6$.

$$D = \sqrt{(4 - (-3))^2 + (-6 - 2)^2}$$ $$D = \sqrt{(4 + 3)^2 + (-8)^2}$$ $$D = \sqrt{(7)^2 + (-8)^2}$$ $$D = \sqrt{49 + 64}$$ $$D = \sqrt{113}$$

$$D \approx 10.6301$$

The distance is:

$$\boxed{10.63 \text{ units}}$$

1. (b) Find the acute angle between the lines $2x + 3y = 5$ and $3x + 2y = 1$ correct to one decimal place.

For the first line, $2x + 3y = 5$:

$$3y = -2x + 5$$ $$y = -\frac{2}{3}x + \frac{5}{3}$$

So, the slope $m_1 = -\frac{2}{3}$.

For the second line, $3x + 2y = 1$:

$$2y = -3x + 1$$ $$y = -\frac{3}{2}x + \frac{1}{2}$$

So, the slope $m_2 = -\frac{3}{2}$.

The acute angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by:

$$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$

Substitute $m_1 = -\frac{2}{3}$ and $m_2 = -\frac{3}{2}$:

$$\tan \theta = \left| \frac{-\frac{2}{3} - \left(-\frac{3}{2}\right)}{1 + \left(-\frac{2}{3}\right)\left(-\frac{3}{2}\right)} \right|$$ $$\tan \theta = \left| \frac{-\frac{2}{3} + \frac{3}{2}}{1 + \frac{6}{6}} \right|$$ $$\tan \theta = \left| \frac{-\frac{4}{6} + \frac{9}{6}}{1 + 1} \right|$$ $$\tan \theta = \left| \frac{\frac{5}{6}}{2} \right|$$ $$\tan \theta = \frac{5}{6} \times \frac{1}{2}$$ $$\tan \theta = \frac{5}{12}$$

$$\theta = \arctan\left(\frac{5}{12}\right)$$ $$\theta \approx 22.61986^\circ$$

Round to one decimal place:

$$\theta = \boxed{22.6^\circ}$$