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.

### Problem 5: Equation of a line through (-4,-2) forming a triangle with an area of 16 units² with the coordinate axes

The general equation of a line is:
\[
\frac{x}{a} + \frac{y}{b} = 1
\]
where \(a\) and \(b\) are the x- and y-intercepts of the line.

The area of the triangle formed by the line with the coordinate axes is given by:
\[
\text{Area} = \frac{1}{2} \times |a| \times |b|
\]
Given the area is 16, we have:
\[
\frac{1}{2} \times |a| \times |b| = 16
\]
\[
|a| \times |b| = 32 \tag{1}
\]

The line passes through the point (-4, -2), so we can substitute \((x, y) = (-4, -2)\) into the line equation:
\[
\frac{-4}{a} + \frac{-2}{b} = 1
\]
\[
\frac{4}{a} + \frac{2}{b} = 1 \tag{2}
\]

Now, solve equations (1) and (2) simultaneously.

From equation (2), we can express one variable in terms of the other. Let's solve for \( \frac{2}{b} \):
\[
\frac{2}{b} = 1 - \frac{4}{a}
\]
\[
b = \frac{2a}{a - 4} \tag{3}
\]

Substitute equation (3) into equation (1):
\[
a \times \frac{2a}{a - 4} = 32
\]
\[
\frac{2a^2}{a - 4} = 32
\]
Multiply both sides by \(a - 4\):
\[
2a^2 = 32(a - 4)
\]
\[
2a^2 = 32a - 128
\]
\[
a^2 - 16a + 64 = 0
\]
Solving this quadratic equation using the quadratic formula:
\[
a = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(64)}}{2(1)}
\]
\[
a = \frac{16 \pm \sqrt{256 - 256}}{2}
\]
\[
a = \frac{16}{2} = 8
\]

Now that we have \(a = 8\), substitute it back into equation (3) to find \(b\):
\[
b = \frac{2(8)}{8 - 4} = \frac{16}{4} = 4
\]

So, the intercepts are \(a = 8\) and \(b = 4\). Thus, the equation of the line is:
\[
\frac{x}{8} + \frac{y}{4} = 1
\]
or
\[
x + 2y = 8
\]

### Problem 6: Equation of lines parallel to \(3x + 4y - 12 = 0\) at a distance of 3 from the origin

The general equation of a line parallel to \(3x + 4y - 12 = 0\) is:
\[
3x + 4y + c = 0
\]
The distance \(d\) of a line \(Ax + By + C = 0\) from the origin \((0, 0)\) is given by:
\[
d = \frac{|C|}{\sqrt{A^2 + B^2}}
\]
For the given line, \(A = 3\), \(B = 4\), and the distance \(d = 3\). So we have:
\[
3 = \frac{|c|}{\sqrt{3^2 + 4^2}} = \frac{|c|}{5}
\]
\[
|c| = 15
\]

Thus, \(c = \pm 15\).

Therefore, the equations of the two lines parallel to \(3x + 4y - 12 = 0\) and at a distance of 3 units from the origin are:
\[
3x + 4y + 15 = 0 \quad \text{or} \quad 3x + 4y - 15 = 0
\]

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Do you want more details or have any questions?

Here are 5 related questions:
1. How do you determine the equation of a line given two points?
2. What is the formula for calculating the distance from a point to a line?
3. How do you find the area of a triangle formed by a line and the coordinate axes?
4. What are the conditions for two lines to be parallel?
5. How does the value of \(C\) affect the position of a line?

**Tip:** To find the distance between a line and a point, always use the distance formula for a line in standard form \(Ax + By + C = 0\).

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 a distance +3 from the origin.

Learn how to find the equation of a line through (-4,-2) that forms a triangle with the coordinate axes with an area of 16 units², and how to determine the equations of lines parallel to 3x + 4y - 12 = 0 at a distance of 3 units from the origin. This problem covers key concepts in coordinate geometry and is suitable for Grades 9-11.

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