Let's go through these one by one, solving each of the problems step by step.

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

The equation of a line in intercept form is: $\frac{x}{a} + \frac{y}{b} = 1$ where $a$ is the x-intercept and $b$ is the y-intercept.

Given:

• $a = 8$
• $b = -1$

Substitute into the formula: $\frac{x}{8} + \frac{y}{-1} = 1$ Simplify the equation: $\frac{x}{8} - y = 1$ Multiply through by 8 to eliminate the fractions: $x - 8y = 8$ Thus, the equation of the line is: $x - 8y = 8$

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

We start with the given equation: $3x - 8y + 24 = 0$ Move the constant term to the other side: $3x - 8y = -24$ Now, divide through by $-24$ to rewrite in intercept form: $\frac{3x}{-24} + \frac{-8y}{-24} = 1$ Simplify: $\frac{x}{-8} + \frac{y}{3} = 1$ Thus, the intercept form is: $\frac{x}{-8} + \frac{y}{3} = 1$

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 written as: $Ax + By + C = 0$ with $A^2 + B^2 = 1$. We first reduce the equation to normal form.

Given: $x + 6y - 7 = 0$ Move the constant to the other side: $x + 6y = 7$ Now, divide through by the square root of the sum of squares of the coefficients of $x$ and $y$ to normalize: $A = 1, B = 6 \implies \sqrt{A^2 + B^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37}$ Divide by $\sqrt{37}$: $\frac{x}{\sqrt{37}} + \frac{6y}{\sqrt{37}} = \frac{7}{\sqrt{37}}$ Thus, the normal form is: $\frac{x}{\sqrt{37}} + \frac{6y}{\sqrt{37}} = \frac{7}{\sqrt{37}}$

The distance from the origin to the line is given by: $d = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{|7|}{\sqrt{37}} = \frac{7}{\sqrt{37}}$

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

Let the x-intercept be $3b$ and the y-intercept be $b$. The equation of the line in intercept form is: $\frac{x}{3b} + \frac{y}{b} = 1$ Multiply through by $b$ to simplify: $\frac{x}{3} + y = b$ Substitute the point $(5, 2)$ into the equation: $\frac{5}{3} + 2 = b$ Simplify: $\frac{5}{3} + \frac{6}{3} = b \implies b = \frac{11}{3}$ Now substitute $b = \frac{11}{3}$ into the original equation: $\frac{x}{3} + y = \frac{11}{3}$ Multiply through by 3: $x + 3y = 11$ Thus, the equation of the line is: $x + 3y = 11$

5. Find the equation of a line through $(-4, -2)$ and forming with the coordinate axes a triangle with area 16.

The area of a triangle formed by a line with the x- and y-intercepts is given by: $\text{Area} = \frac{1}{2} \times |x\text{-intercept}| \times |y\text{-intercept}|$ Let the x-intercept be $a$ and the y-intercept be $b$. The equation of the line in intercept form is: $\frac{x}{a} + \frac{y}{b} = 1$ The area is given as 16, so: $\frac{1}{2} \times |a| \times |b| = 16$ $|a| \times |b| = 32$ Also, the line passes through the point $(-4, -2)$, so substituting into the