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 $Ax + By + C = 0$.