Question

In the given figure, point D divides the side BC of $\Delta ABC$ in the ratio $1 : 2$. Find length AD.

Hint:

Double check that the ratio order ($BD : DC = 1 : 2$) corresponds correctly to your section formula variables: $m_1$ multiplies $x_2$ (the coordinate of $C$), and $m_2$ multiplies $x_1$ (the coordinate of $B$).

Answer 1

Step 1: Find point D using the section formula.
$D$ divides $BC$ in ratio $1:2$ where $B(-2,1)$ and $C(4,2)$:
\[ D = \left(\frac{1(4)+2(-2)}{3}, \frac{1(2)+2(1)}{3}\right) = \left(0, \frac{4}{3}\right) \]
Step 2: Instead of the distance formula, find $AD$ using the horizontal and vertical legs of a right triangle.
The horizontal gap between $A(1,5)$ and $D(0,\frac43)$ is $1$ unit, and the vertical gap is $5-\frac43=\frac{11}{3}$ units.
Step 3: Apply Pythagoras' theorem on these two legs.
\[ AD = \sqrt{1^2 + \left(\frac{11}{3}\right)^2} = \sqrt{1 + \frac{121}{9}} = \sqrt{\frac{130}{9}} \]
Step 4: Simplify and conclude.
\[ AD = \frac{\sqrt{130}}{3}\text{ units} \]
\[ \boxed{AD = \frac{\sqrt{130}}{3}\text{ units}} \]