Correct Option: A To determine the equation of the line, we can use the intercept form of a straight line, which is $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept. Given that the line forms a triangle of area $24$ square units with the coordinate axes in the first quadrant, it implies that both intercepts $a$ and $b$ are positive ($a > 0, b > 0$). The vertices of this triangle are $(0,0)$, $(a,0)$, and $(0,b)$. The area of a triangle with vertices at the origin and intercepts $(a,0)$ and $(0,b)$ is given by the formula $\frac{1}{2} \times |a| \times |b|$. Since $a,b > 0$, the area is $\frac{1}{2}ab$. We are given that the area is $24$ square units: $\frac{1}{2}ab = 24$ $ab = 48 \quad \cdots (1)$ Next, the problem states that the line passes through the point $(3, 4)$. We can substitute these coordinates into the intercept form of the line's equation: $\frac{3}{a} + \frac{4}{b} = 1 \quad \cdots (2)$ Now we have a system of two equations with two variables, $a$ and $b$. From equation (1), we can express $b$ in terms of $a$: $b = \frac{48}{a}$. Substitute this expression for $b$ into equation (2): $\frac{3}{a} + \frac{4}{(48/a)} = 1$ $\frac{3}{a} + \frac{4a}{48} = 1$ Simplify the second term: $\frac{3}{a} + \frac{a}{12} = 1$ To solve for $a$, we can eliminate the denominators by multiplying the entire equation by the least common multiple of $a$ and $12$, which is $12a$: $12a \left( \frac{3}{a} \right) + 12a \left( \frac{a}{12} \right) = 12a \times 1$ $36 + a^2 = 12a$ Rearrange the terms to form a standard quadratic equation: $a^2 - 12a + 36 = 0$ This quadratic equation is a perfect square trinomial, which can be factored as: $(a - 6)^2 = 0$ Solving for $a$: $a = 6$ Now that we have the value of $a$, we can substitute it back into equation (1) to find $b$: $6b = 48$ $b = \frac{48}{6}$ $b = 8$ So, the x-intercept is $a=6$ and the y-intercept is $b=8$. Substitute these values back into the intercept form of the line's equation: $\frac{x}{6} + \frac{y}{8} = 1$ To convert this equation to the standard form $Ax + By = C$, we multiply the entire equation by the least common multiple (LCM) of the denominators $6$ and $8$, which is $24$: $24 \left( \frac{x}{6} \right) + 24 \left( \frac{y}{8} \right) = 24 \times 1$ $4x + 3y = 24$ This matches option A. To verify the solution: 1. Check if $(3,4)$ lies on $4x + 3y = 24$: $4(3) + 3(4) = 12 + 12 = 24$. (Correct) 2. Check the area with intercepts: If $x=0$, $3y=24 \Rightarrow y=8$. If $y=0$, $4x=24 \Rightarrow x=6$. The x-intercept is $6$ and y-intercept is $8$. The area is $\frac{1}{2} \times 6 \times 8 = 24$. (Correct) The final answer is $\boxed{\text{A}}$