If a line intercepted between the coordinate axes is trisected at a point A(4, 3), which is nearer to x-axis, then its equation is:

We are given that a line segment intercepted between the coordinate axes is trisected at the point A(4, 3), and A is nearer to the x-axis. We need to find the equation of this line.

Let the line intersect the x-axis at point P(a, 0) and the y-axis at point Q(0, b). The segment PQ is trisected, meaning it is divided into three equal parts by two points. Since A(4, 3) is nearer to the x-axis, it must be the trisection point closer to P (the x-intercept). Therefore, A divides PQ in the ratio 1:2, where PA:AQ = 1:2 (the part from P to A is one-third of PQ).

Using the section formula, if a point divides the line segment joining (x₁, y₁) and (x₂, y₂) in the ratio m:n, its coordinates are:

$$ \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) $$

Here, P(a, 0) and Q(0, b), and A(4, 3) divides PQ in the ratio m:n = 1:2. Substituting:

For the x-coordinate:

$$ 4 = \frac{1 \cdot 0 + 2 \cdot a}{1 + 2} = \frac{2a}{3} $$

Solving for a:

$$ 4 = \frac{2a}{3} $$

Multiply both sides by 3:

$$ 12 = 2a $$

Divide both sides by 2:

$$ a = 6 $$

For the y-coordinate:

$$ 3 = \frac{1 \cdot b + 2 \cdot 0}{1 + 2} = \frac{b}{3} $$

Solving for b:

$$ 3 = \frac{b}{3} $$

Multiply both sides by 3:

$$ b = 9 $$

So, the x-intercept is 6 and the y-intercept is 9. The equation of a line in intercept form is:

$$ \frac{x}{a} + \frac{y}{b} = 1 $$

Substituting a = 6 and b = 9:

$$ \frac{x}{6} + \frac{y}{9} = 1 $$

To eliminate denominators, multiply both sides by the least common multiple of 6 and 9, which is 18:

$$ 18 \cdot \frac{x}{6} + 18 \cdot \frac{y}{9} = 18 \cdot 1 $$

Simplify each term:

$$ 3x + 2y = 18 $$

Thus, the equation is 3x + 2y = 18.

Now, comparing with the options:

A. 4x - 3y = 7

B. 3x + 2y = 18

C. 3x + 8y = 36

D. x + 3y = 13

Option B matches our equation. To verify, check if A(4, 3) lies on this line and if it trisects PQ with intercepts (6, 0) and (0, 9).

Substitute x = 4, y = 3 into 3x + 2y:

$$ 3(4) + 2(3) = 12 + 6 = 18 $$

This equals the right-hand side, so A lies on the line.

For trisection: the point dividing PQ from P(6, 0) to Q(0, 9) in the ratio 1:2 is:

$$ \left( \frac{1 \cdot 0 + 2 \cdot 6}{1 + 2}, \frac{1 \cdot 9 + 2 \cdot 0}{1 + 2} \right) = \left( \frac{12}{3}, \frac{9}{3} \right) = (4, 3) $$

This confirms A(4, 3) is the trisection point nearer to the x-axis (since the other trisection point at ratio 2:1 is (2, 6), which has a larger y-coordinate and is farther from the x-axis).

Hence, the correct answer is Option B.