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: Understanding the Question:
We are given a triangle $\Delta ABC$ with vertices $A(1, 5)$, $B(-2, 1)$, and $C(4, 2)$.
A point $D$ lies on the side $BC$ such that it divides $BC$ internally in the ratio $1 : 2$ ($BD : DC = 1 : 2$).
We need to find the coordinates of $D$ first, and then calculate the distance $AD$.

Step 2: Key Formula or Approach:
- The section formula to find coordinates of $D(x, y)$ dividing segment $BC$ in ratio $m_1 : m_2$ is:
\[ x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \quad y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \]
- The distance formula to find the length $AD$ is:
\[ AD = \sqrt{(x_D - x_A)^2 + (y_D - y_A)^2} \]

Step 3: Detailed Explanation:

• Identify the coordinates of the endpoints:
\[ B(x_1, y_1) = (-2, 1) \]
\[ C(x_2, y_2) = (4, 2) \]
The ratio of division is $m_1 : m_2 = 1 : 2$.

• Apply the section formula to find coordinates of $D(x, y)$:
\[ x = \frac{1(4) + 2(-2)}{1+2} = \frac{4 - 4}{3} = 0 \]
\[ y = \frac{1(2) + 2(1)}{1+2} = \frac{2 + 2}{3} = \frac{4}{3} \]
Therefore, the coordinates of point $D$ are $\left(0, \frac{4}{3}\right)$.

• Use the distance formula to find the length of $AD$ between $A(1, 5)$ and $D\left(0, \frac{4}{3}\right)$:
\[ AD = \sqrt{(0 - 1)^2 + \left(\frac{4}{3} - 5\right)^2} \]
\[ AD = \sqrt{(-1)^2 + \left(\frac{4 - 15}{3}\right)^2} \]
\[ AD = \sqrt{1 + \left(-\frac{11}{3}\right)^2} \]
\[ AD = \sqrt{1 + \frac{121}{9}} \]
\[ AD = \sqrt{\frac{9 + 121}{9}} = \sqrt{\frac{130}{9}} = \frac{\sqrt{130}}{3}\text{ units} \]


Step 4: Final Answer:
The length of segment $AD$ is $\frac{\sqrt{130}}{3}$ units.