The line segment AB meets the X-axis at A and the Y-axis at B. The point P(-3, 4) on AB divides it in the ratio 2:3. What are the coordinates of points A and B, respectively?

Let the coordinates of point A be (x, 0) and the coordinates of point B be (0, y).

Using the section formula, we can find the coordinates of point P:
x = (2*(-3) + 3*0)/(2+3) = -6/5
y = (2*0 + 3*4)/(2+3) = 12/5

Therefore, the coordinates of point P are (-6/5, 12/5).

Since point P lies on the line segment AB, we can write the following equations:
(12/5 - 0)/(0 - x) = 3/2
(0 - (-6/5))/(y - 4) = 2/3

Solving these equations, we get:
x = -6/5
y = 12/5

Therefore, the coordinates of points A and B are (-6/5, 0) and (0, 12/5) respectively.

Let A be (-a, 0) and B be (0, b).

Since point P divides AB in the ratio 2:3, we have:

$\frac{AP}{PB} = \frac{2}{3}$

Using the distance formula, we can find the lengths of AP and PB:

$AP = \sqrt{(-a+3)^2+(4-4)^2} = \sqrt{a^2-6a+9}$

$PB = \sqrt{(0+3)^2+(b-4)^2} = \sqrt{9+(b-4)^2}$

Substituting the lengths into the ratio equation, we get:

$\frac{\sqrt{a^2-6a+9}}{\sqrt{9+(b-4)^2}} = \frac{2}{3}$

Squaring both sides:

$\frac{a^2-6a+9}{9+(b-4)^2} = \frac{4}{9}$

Cross-multiplying:

$9(a^2-6a+9) = 4[9+(b-4)^2]$

Simplifying:

$a^2-6a+81 = 36+4(b^2-8b+16)$

$a^2-6a+45 = 4b^2-32b+40$

$a^2-6a-4b^2+32b+5 = 0$

Now, we can use the fact that A lies on the X-axis and B lies on the Y-axis to solve for a and b.

Since A lies on the X-axis, its y-coordinate is 0. Substituting this into the equation, we get:

$a^2-6a-4(0)^2+32(0)+5 = 0$

$a^2-6a+5 = 0$

This is a quadratic equation that can be solved using the quadratic formula:

$a = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$a = \frac{6 \pm \sqrt{(-6)^2-4(1)(5)}}{2(1)}$

$a = \frac{6 \pm \sqrt{36-20}}{2}$

$a = \frac{6 \pm \sqrt{16}}{2}$

$a = 3 \pm 2$

So, either $a = 1$ or $a = 5$. Since $a$ is the x-coordinate of point A, we have:

$A = (-1, 0)$ or $A = (-5, 0)$

Since B lies on the Y-axis, its x-coordinate is 0. Substituting this into the equation, we get:

$(0)^2-6(0)-4b^2+32b+5 = 0$

$-4b^2+32b+5 = 0$

This is a quadratic equation that can be solved using the quadratic formula:

$b = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$b = \frac{-32 \pm \sqrt{(-32)^2-4(-4)(5)}}{2(-4)}$

$b = \frac{-32 \pm \sqrt{1024+80}}{-8}$

$b = \frac{-32 \pm \sqrt{1104}}{-8}$

$b = \frac{-32 \pm 33.23}{-8}$

So, either $b = \frac{1.23}{8}$ or $b = 6.53$. Since $b$ is the y-coordinate of point B, we have:

$B = (0, \frac{1.23}{8})$ or $B = (0, 6.53)$

Therefore, the coordinates of points A and B are $(-1, 0)$ and $(0, \frac{1.23}{8})$, or $(-5, 0)$ and $(0, 6.53)$.