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: Find Intercepts
For x-intercept (A), put $y=0$: $3x = 24 \implies x=8$. So, $A(8, 0)$.
For y-intercept (B), put $x=0$: $4y = 24 \implies y=6$. So, $B(0, 6)$.

Step 2: Triangle Dimensions
The triangle is right-angled at $O(0,0)$. Length of sides: $a = OB = 6$, $b = OA = 8$.
Hypotenuse $c = AB = \sqrt{6^2 + 8^2} = 10$.

Step 3: Incentre Formula
For a right-angled triangle with legs $a, b$ and hypotenuse $c$, the inradius $r = \frac{a+b-c}{2}$.
$r = \frac{6+8-10}{2} = \frac{4}{2} = 2$.

Step 4: Conclusion
Since the triangle is in the first quadrant and the sides are the axes, the incentre $(I)$ is $(r, r) = (2, 2)$.
Final Answer: (B)