Question

If the line $3x + 4y - 24 = 0$ intersects X and Y axes in points A and B respectively then incentre of the triangle OAB where O is origin is

• A. (4, 4)
• B. (2, 2)
• C. (3, 4)
• D. (4, 3)

Hint:

For any triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and opposite sides $a, b, c$, Incentre $I = \left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}\right)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the incenter of a right-angled triangle formed by the origin and the axes intercepts of a given line.
Step 2: Key Formula or Approach:
For a right-angled triangle with legs of length \( a \) and \( b \) and hypotenuse \( c \), the inradius \( r = \frac{a + b - c}{2} \).
For such a triangle with vertices at \( (0,0), (a,0), (0,b) \), the incenter is \( (r, r) \).
Step 3: Detailed Explanation:
1. Find intercepts of \( 3x + 4y = 24 \).
- X-intercept (put \( y=0 \)): \( 3x = 24 \implies x = 8 \). Point \( A = (8, 0) \).
- Y-intercept (put \( x=0 \)): \( 4y = 24 \implies y = 6 \). Point \( B = (0, 6) \).
2. Triangle vertices are \( O(0,0), A(8,0), B(0,6) \).
3. Leg lengths are \( OA = 8 \) and \( OB = 6 \).
4. Hypotenuse \( AB = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10 \).
5. Inradius \( r = \frac{8 + 6 - 10}{2} = \frac{4}{2} = 2 \).
6. Since the triangle is in the first quadrant, the incenter is \( (r, r) = (2, 2) \).
Step 4: Final Answer:
The incenter is \( (2, 2) \).