Question

From the figure given alongside, find the length of the median AD of triangle ABC. Complete the activity.

Solution:

Here A(–1, 1), B(5, – 3), C(3, 5) and suppose D(x, y) are coordinates of point D.

Using midpoint formula,

x = ${\displaystyle \frac{5+3}{2}}$

∴ x = ${\displaystyle \square }$

y = ${\displaystyle \frac{-3+5}{2}}$

∴ y = ${\displaystyle \square }$

Using distance formula,

∴ AD = ${\displaystyle \sqrt{{(4-\square )}^2+{(1-1)}^2}}$

∴ AD = ${\displaystyle \sqrt{{(\square )}^2+{\left(0\right)}^2}}$

∴ AD = ${\displaystyle \sqrt{\square }}$

∴ The length of median AD = ${\displaystyle \square }$

Answer 1

Here A(–1, 1), B(5, – 3), C(3, 5) and suppose D(x, y) are coordinates of point D. D is the midpoint of seg BC.

Using midpoint formula,

x = ${\displaystyle \frac{x_1+x_2}{2}}$

x = ${\displaystyle \frac{5+3}{2}}$

∴ x = ${\displaystyle \frac{8}{2}}$

∴ x = 4

y = ${\displaystyle \frac{y_1+y_2}{2}}$

y = ${\displaystyle \frac{-3+5}{2}}$

∴ y = ${\displaystyle \frac{2}{2}}$

∴ y = 1

Using distance formula,

∴ AD = ${\displaystyle \sqrt{{(4-(-1))}^2+{(1-1)}^2}}$

∴ AD = ${\displaystyle \sqrt{{\left(5\right)}^2+{\left(0\right)}^2}}$

∴ AD = ${\displaystyle \sqrt{25}}$

∴ The length of median AD = 5 cm