Let's go through the problems one by one and solve them with detailed explanations.

1. Find the equation of the line whose x-intercept is 8 and y-intercept is -1.

To solve this, use the intercept form of the equation of a line:

$\frac{x}{a} + \frac{y}{b} = 1$

where $a$ is the x-intercept and $b$ is the y-intercept.

• Here, $a = 8$ and $b = -1$.

Substitute the values:

$\frac{x}{8} + \frac{y}{-1} = 1$

Multiply through by 8 to eliminate the denominators:

$x - 8y = 8$

This is the required equation of the line.

2. Reduce $3x - 8y + 24 = 0$ to the intercept form.

The intercept form is:

$\frac{x}{a} + \frac{y}{b} = 1$

We need to manipulate the given equation to fit this form.

Start by moving the constant term to the other side:

$3x - 8y = -24$

Now, divide the entire equation by $-24$ (the constant term on the right-hand side):

$\frac{3x}{-24} - \frac{8y}{-24} = 1$

Simplify the fractions:

$\frac{x}{-8} + \frac{y}{3} = 1$

Thus, the equation in intercept form is:

$\frac{x}{-8} + \frac{y}{3} = 1$

where the x-intercept is $-8$ and the y-intercept is $3$.

3. Reduce $x + 6y - 7 = 0$ to the normal form. Give the distance of the line from the origin.

The normal form of a line is:

$x \cos \theta + y \sin \theta = p$

where $p$ is the perpendicular distance from the origin to the line and $\theta$ is the angle the normal makes with the x-axis.

We begin by rewriting the given equation:

$x + 6y = 7$

Now, express it in the form:

$Ax + By = C$

where $A = 1$, $B = 6$, and $C = 7$.

To reduce this to normal form, divide the equation by the length of the normal (which is the square root of $A^2 + B^2$):

$\text{Length of the normal} = \sqrt{1^2 + 6^2} = \sqrt{37}$

Dividing the equation by $\sqrt{37}$:

$\frac{x}{\sqrt{37}} + \frac{6y}{\sqrt{37}} = \frac{7}{\sqrt{37}}$

This is the normal form of the line. The distance of the line from the origin is the constant term on the right-hand side:

$\text{Distance} = \frac{7}{\sqrt{37}} \approx 1.15$

4. What is the equation of a line through $(5, 2)$ with the x-intercept thrice the y-intercept?

Let the x-intercept be $a$ and the y-intercept be $b$. We are told that the x-intercept is thrice the y-intercept, so:

$a = 3b$

Using the intercept form of the line equation:

$\frac{x}{a} + \frac{y}{b} = 1$

Substitute $a = 3b$:

$\frac{x}{3b} + \frac{y}{b} = 1$

Now, multiply through by (b\