Question

Let A \( \equiv \) (0,0), B(3, 0), C(0, -4) are vertices of \( \triangle \)ABC, then the co-ordinates of incentre of \( \triangle \)ABC is

• A. \( \left( \frac{45}{14}, \frac{3}{14} \right) \)
• B. \( \left( \frac{45}{14}, \frac{45}{14} \right) \)
• C. \( \left( \frac{3}{14}, \frac{45}{14} \right) \)
• D. \( \left( \frac{45}{14}, -\frac{45}{14} \right) \)

Hint:

Incentre $(I) = \frac{\sum a x_i}{\sum a}$.

Answer 1

The Correct Option is A

Step 1: Find side lengths
$c$ (side opposite C) $= AB = 3$
$b$ (side opposite B) $= AC = 4$
$a$ (side opposite A) $= BC = \sqrt{3^2 + (-4)^2} = 5$.
Step 2: Incentre formula
$I = \left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c} \right)$.
$a+b+c = 5+4+3 = 12$.
Step 3: Calculation
$x = \frac{5(0) + 4(3) + 3(0)}{12} = \frac{12}{12} = 1$.
$y = \frac{5(0) + 4(0) + 3(-4)}{12} = \frac{-12}{12} = -1$.
Step 4: Conclusion
The coordinates are (1, -1). (Note: Re-evaluate original prompt options for closest match or typos).
Final Answer:(None/Manual)