Question

In the given figure, point D divides the side BC of \(\triangle ABC\) in the ratio \(1 : 2\). Find length AD. (Given coordinates: \(A(1, 5), B(-2, 1), C(4, 2)\))

Hint:

Always double-check the order of \(m\) and \(n\) in the section formula relative to points \(B\) and \(C\). Since the ratio is \(1:2\) from \(B\) to \(C\), \(m=1\) multiplies the coordinates of \(C\).

Answer 1

We are given:
A(1, 5), B(–2, 1), C(4, 2)
D divides BC in the ratio 1 : 2.

Step 1: Find coordinates of D using section formula
D divides BC in ratio 1 : 2 → BD : DC = 1 : 2.

Coordinates of D:
\[ D(x, y) = \left( \frac{1 \cdot 4 + 2 \cdot (-2)}{1 + 2},\ \frac{1 \cdot 2 + 2 \cdot 1}{1 + 2} \right) \] Simplify x–coordinate:
\[ x = \frac{4 - 4}{3} = 0 \] Simplify y–coordinate:
\[ y = \frac{2 + 2}{3} = \frac{4}{3} \]
So, \[ D\left(0,\ \frac{4}{3}\right) \]

Step 2: Find length AD using distance formula
A(1, 5), D\(\left(0,\frac{4}{3}\right)\)

\[ AD = \sqrt{(1 - 0)^2 + \left(5 - \frac{4}{3}\right)^2} \] \[ = \sqrt{1^2 + \left(\frac{15}{3} - \frac{4}{3}\right)^2} \] \[ = \sqrt{1 + \left(\frac{11}{3}\right)^2} \] \[ = \sqrt{1 + \frac{121}{9}} \] \[ = \sqrt{\frac{9 + 121}{9}} \] \[ = \sqrt{\frac{130}{9}} \] \[ = \frac{\sqrt{130}}{3} \]

Final Answer:
Coordinates of D: \[ \left(0,\ \frac{4}{3}\right) \]
Length AD: \[ \boxed{\frac{\sqrt{130}}{3}} \]