`A(4,3),B(6,5)and C(5,-2)` are the vertices of a `DeltaABC,` if P is a point on BC such that `BP:PC =2:3.` Find the co-ordinates of P and then prove then ar `(DeltaABP):ar(DeltaACP)=2:3.`

Answer

To find the coordinates of point P on line segment BC such that the ratio \( BP:PC = 2:3 \), we can use the section formula for internal division.

### Step 1: Identify the coordinates of points B and C
- Point B has coordinates \( B(6, 5) \)
- Point C has coordinates \( C(5, -2) \)

### Step 2: Use the section formula
The section formula states that if a point P divides the line segment joining points \( (x_1, y_1) \) and \( (x_2, y_2) \) in the ratio \( m:n \), then the coordinates of point P are given by:

\[ P\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \]

Here, \( m = 2 \) and \( n = 3 \), and we have:
- \( (x_1, y_1) = (6, 5) \) (coordinates of B)
- \( (x_2, y_2) = (5, -2) \) (coordinates of C)

### Step 3: Calculate the x-coordinate of P
Using the section formula for the x-coordinate:

\[ x_P = \frac{2 \cdot 5 + 3 \cdot 6}{2 + 3} = \frac{10 + 18}{5} = \frac{28}{5} = 5.6 \]

### Step 4: Calculate the y-coordinate of P
Using the section formula for the y-coordinate:

\[ y_P = \frac{2 \cdot (-2) + 3 \cdot 5}{2 + 3} = \frac{-4 + 15}{5} = \frac{11}{5} = 2.2 \]

### Step 5: Write the coordinates of P
Thus, the coordinates of point P are \( P\left( \frac{28}{5}, \frac{11}{5} \right) \) or \( P(5.6, 2.2) \).

### Step 6: Prove the area ratio
To prove that the area of triangle \( ABP \) to triangle \( ACP \) is \( 2:3 \), we note that both triangles share a common height from point A to line BC. Therefore, the ratio of their areas is equal to the ratio of their bases, which are segments BP and PC.

Given that \( BP:PC = 2:3 \), we can conclude:

\[ \text{Area}(ABP) : \text{Area}(ACP) = BP : PC = 2 : 3 \]

### Final Answer
The coordinates of point P are \( P(5.6, 2.2) \) and the area ratio \( \text{Area}(ABP) : \text{Area}(ACP) = 2 : 3 \).

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