If A = (-4, 3) and B = (8, -6),

(i) find the length of AB.

(ii) in what ratio is the line segment joining AB, divided by the x-axis?

The graph is given below:

(i) Given,

A = (-4, 3), B = (8, -6)

$\text{Length of AB = } \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] = \sqrt{(8 - (-4))^2 + (-6 - 3)^2} \\[1em] = \sqrt{(8 + 4)^2 + (-6 - 3)^2} \\[1em] = \sqrt{12^2 + (-9)^2} \\[1em] = \sqrt{144 + 81} \\[1em] = \sqrt{225} \\[1em] = 15.

Hence, the length of AB = 15 units.

(ii) From graph, we see that O (0, 0) lies on AB.
Let O divide AB in the ratio m₁ : m₂.

By section-formula the coordinates of point dividing a line in m₁ : m₂ are given by

$\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Putting values in above equation for finding x coordinates we get,

$\Rightarrow \dfrac{m$1 \times 8 + m2 \times -4}{m1 + m2} \ \\[1em] \Rightarrow \dfrac{8m1 - 4m2}{m1 + m2}.

Since origin (0, 0) is the dividing point the x-coordinate = 0. Comparing it with above equation we get,

$\dfrac{8m$1 - 4m2}{m1 + m2} = 0 \\[1em] 8m1 - 4m2 = 0 \\[1em] 8m1 = 4m2 \\[1em] \dfrac{m1}{m2} = \dfrac{4}{8} \\[1em] \dfrac{m1}{m2} = \dfrac{1}{2}.

∴ m₁ : m₂ = 1 : 2.

Hence, the x-axis divides the line segment AB in the ratio 1 : 2.