Question:
If the line \(3x+4y-24=0\) intersects the x-axis at the point \(A\) and the y-axis at the point \(B\), then the incentre of the triangle \(OAB\), where \(O\) is the origin, is :
If the line \(3x+4y-24=0\) intersects the x-axis at the point \(A\) and the y-axis at the point \(B\), then the incentre of the triangle \(OAB\), where \(O\) is the origin, is :
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Find A and B, then use the incentre formula weighted by the opposite side lengths.
Updated On: Nov 5, 2026
- (3, 4)
- (2, 2)
- (4, 4)
- (4, 3)
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The Correct Option is B
Solution and Explanation
\(A=(8,0)\), \(B=(0,6)\), \(O=(0,0)\). Sides: \(OA=8\), \(OB=6\), \(AB=10\).
Incentre \(=\left(\frac{a x_1+b x_2+c x_3}{a+b+c},\frac{a y_1+b y_2+c y_3}{a+b+c}\right)\) with \(a=AB=10\) (opposite \(O\)), \(b=OB=6\) (opposite \(A\)), \(c=OA=8\) (opposite \(B\)).
\(x=\frac{10\cdot0+6\cdot8+8\cdot0}{24}=2\), \(y=\frac{10\cdot0+6\cdot0+8\cdot6}{24}=2\).
Incentre \(=(2,2)\).
Answer: (2) (2, 2)
Incentre \(=\left(\frac{a x_1+b x_2+c x_3}{a+b+c},\frac{a y_1+b y_2+c y_3}{a+b+c}\right)\) with \(a=AB=10\) (opposite \(O\)), \(b=OB=6\) (opposite \(A\)), \(c=OA=8\) (opposite \(B\)).
\(x=\frac{10\cdot0+6\cdot8+8\cdot0}{24}=2\), \(y=\frac{10\cdot0+6\cdot0+8\cdot6}{24}=2\).
Incentre \(=(2,2)\).
Answer: (2) (2, 2)
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