Question

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

• A. \((1,-1)\)
• B. \((\frac{45}{14},\frac{-3}{14})\)
• C. \((\frac{3}{14},\frac{45}{14})\)
• D. \((\frac{-3}{14},\frac{45}{14})\)

Hint:

Use the incentre formula with weights equal to the opposite side lengths.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The incentre is \(\left(\dfrac{ax_1+bx_2+cx_3}{a+b+c},\ \dfrac{ay_1+by_2+cy_3}{a+b+c}\right)\), where \(a,b,c\) are the sides opposite \(A,B,C\).

Step 2: Side Lengths
\(a=BC=\sqrt{9+16}=5\), \(b=CA=4\), \(c=AB=3\).

Step 3: Calculation
\(x=\dfrac{5(0)+4(3)+3(0)}{12}=1\)
\(y=\dfrac{5(0)+4(0)+3(-4)}{12}=-1\)

Step 4: Compare the options
The incentre is \((1,-1)\), the first option.

Final Answer:
The incentre is the point (1, -1). \[ \boxed{\text{(A) }(1,-1)} \]