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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

Solve the following problems. Find the equation of the line whose x-intercept is 8 and y-intercept is -1, and other related linear algebra questions.

Learn how to solve linear equations involving intercept form, normal form, and calculating distances and triangle areas. Detailed solutions for each problem from Exercise 4.3. Suitable for high school students (Grades 9-12).

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