The origin O,B (-6,9) and C (12, -3) are vertices of triangle OBC, Point P divides OB in the ratio 1 : 2 and point Q divides OC in the ratio 1 : 2 Find the co-ordinates of points P and Q. Also show that `PQ = (1)/(3)` BC.

Answer

To solve the problem step by step, we will find the coordinates of points P and Q, and then calculate the distance PQ and BC to show that \( PQ = \frac{1}{3} BC \).

### Step 1: Find the coordinates of point P
Point P divides OB in the ratio 1:2. We can use the section formula to find the coordinates of P.

The coordinates of points O and B are:
- O (0, 0)
- B (-6, 9)

Using the section formula:
\[ P(x, y) = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \]
where \( m = 1 \), \( n = 2 \), \( O(0, 0) \) is \( (x_1, y_1) \) and \( B(-6, 9) \) is \( (x_2, y_2) \).

Calculating the x-coordinate of P:
\[ x_P = \frac{1 \cdot (-6) + 2 \cdot 0}{1 + 2} = \frac{-6 + 0}{3} = \frac{-6}{3} = -2 \]

Calculating the y-coordinate of P:
\[ y_P = \frac{1 \cdot 9 + 2 \cdot 0}{1 + 2} = \frac{9 + 0}{3} = \frac{9}{3} = 3 \]

Thus, the coordinates of point P are \( P(-2, 3) \).

### Step 2: Find the coordinates of point Q
Point Q divides OC in the ratio 1:2. We will again use the section formula.

The coordinates of points O and C are:
- O (0, 0)
- C (12, -3)

Using the section formula:
\[ Q(x, y) = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \]
where \( m = 1 \), \( n = 2 \), \( O(0, 0) \) is \( (x_1, y_1) \) and \( C(12, -3) \) is \( (x_2, y_2) \).

Calculating the x-coordinate of Q:
\[ x_Q = \frac{1 \cdot 12 + 2 \cdot 0}{1 + 2} = \frac{12 + 0}{3} = \frac{12}{3} = 4 \]

Calculating the y-coordinate of Q:
\[ y_Q = \frac{1 \cdot (-3) + 2 \cdot 0}{1 + 2} = \frac{-3 + 0}{3} = \frac{-3}{3} = -1 \]

Thus, the coordinates of point Q are \( Q(4, -1) \).

### Step 3: Calculate the distance PQ
Using the distance formula:
\[ PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2} \]
Substituting the coordinates of P and Q:
\[ PQ = \sqrt{(4 - (-2))^2 + (-1 - 3)^2} = \sqrt{(4 + 2)^2 + (-4)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} \]

### Step 4: Calculate the distance BC
Using the distance formula:
\[ BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} \]
Substituting the coordinates of B and C:
\[ BC = \sqrt{(12 - (-6))^2 + (-3 - 9)^2} = \sqrt{(12 + 6)^2 + (-12)^2} = \sqrt{18^2 + (-12)^2} = \sqrt{324 + 144} = \sqrt{468} \]

### Step 5: Show that \( PQ = \frac{1}{3} BC \)
To show that \( PQ = \frac{1}{3} BC \):
\[ \frac{PQ}{BC} = \frac{\sqrt{52}}{\sqrt{468}} = \sqrt{\frac{52}{468}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \]

Thus, we have shown that \( PQ = \frac{1}{3} BC \).

### Summary of Results
- Coordinates of P: \( (-2, 3) \)
- Coordinates of Q: \( (4, -1) \)
- \( PQ = \sqrt{52} \)
- \( BC = \sqrt{468} \)
- \( PQ = \frac{1}{3} BC \)