A(10, 5), B(6, -3) and C(2, 1) are the vertices of a ΔABC. L is the mid-point of AB and M is the mid-point of AC. Write down the co-ordinates of L and M. Show that $LM = \dfrac{1}{2} BC.$

By using mid-point formula,

(x, y) = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

L is the mid-point of AB.

Substituting values we get :

$\Rightarrow L = \Big(\dfrac{10 + 6}{2}, \dfrac{5 + (-3)}{2}\Big) \\[1em] \Rightarrow L = \Big(\dfrac{16}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow L = (8, 1).$

Given,

M is the mid-point of AC.

Substituting values we get :

$\Rightarrow M = \Big(\dfrac{10 + 2}{2}, \dfrac{5 + 1}{2}\Big) \\[1em] \Rightarrow M = \Big(\dfrac{12}{2}, \dfrac{6}{2}\Big) \\[1em] \Rightarrow M = (6, 3).$

By distance formula,

d = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Substituting values we get :

$LM = \sqrt{(6 - 8)^2 + (3 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} = 2\sqrt{2}. \\[1em] BC = \sqrt{(2 - 6)^2 + (1 - (-3))^2} \\[1em] = \sqrt{(-4)^2 + (4)^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32} \\[1em] = \sqrt{16 \times 2} \\[1em] = 4\sqrt{2} \\[1em] = 2 \times 2\sqrt{2} \\[1em] = 2 \times LM.$

Thus, BC = 2LM or LM = $\dfrac{1}{2}BC.$

Hence, proved that $LM = \dfrac{1}{2}BC$.