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

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### **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:

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

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### **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\

Solve the following problems:
1. Find the equation of the line whose x-intercept is 8 and y-intercept is -1.
2. Reduce 3x - 8y + 24 = 0 to the intercept form.
3. Reduce x + 6y - 7 = 0 to the normal form. Give the distance of the line from the origin.
4. What is the equation of a line through (5, 2) with the x-intercept thrice the y-intercept?
5. Find the equation of a line through (-4, -2) and forming with the coordinate axes a triangle with area 16.
6. Find the equations of the lines parallel to the line 3x + 4y - 12 = 0 and passing at distance ±3 from the origin.

This exercise covers a variety of coordinate geometry problems, including finding the equation of lines with specific intercepts, reducing equations to intercept and normal form, calculating distances, and determining equations for lines forming triangles with given areas. Suitable for high school students in Grades 10-12.

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