Question

\(\mathrm{A}(10, 5), \mathrm{B}(6, -3)\) and \(\mathrm{C}(2, 1)\) are the vertices of a \(\triangle \mathrm{ABC}\). \(\mathrm{L}\) 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 \(\mathrm{LM} = \frac{1}{2} \mathrm{BC}\).

Answer 1

Co-ordinates of L (mid-point of AB):

$$L = \left(\frac{10 + 6}{2}, \frac{5 + (-3)}{2}\right)$$

$$= \left(\frac{16}{2}, \frac{2}{2}\right)$$

$$= (8, 1)$$

Co-ordinates of M (mid-point of AC):

$$M = \left(\frac{10 + 2}{2}, \frac{5 + 1}{2}\right)$$

$$= \left(\frac{12}{2}, \frac{6}{2}\right)$$

$$= (6, 3)$$

Length LM:

$$LM = \sqrt{(8 - 6)^2 + (1 - 3)^2}$$

$$= \sqrt{4 + 4}$$

$$= \sqrt{8}$$

$$= 2\sqrt{2}\text{ units}$$

Length BC:

$$BC = \sqrt{(6 - 2)^2 + (-3 - 1)^2}$$

$$= \sqrt{16 + 16}$$

$$= \sqrt{32}$$

$$= 4\sqrt{2}\text{ units}$$

Since $$2\sqrt{2} = \dfrac{1}{2}(4\sqrt{2})$$, we get $${LM = \frac{1}{2}BC}$$