Question:
The Incenter of the triangle whose vertices are \(A(0,3,0)\), \(B(0,0,4)\), and \(C(0,3,4)\) is;
The Incenter of the triangle whose vertices are \(A(0,3,0)\), \(B(0,0,4)\), and \(C(0,3,4)\) is;
Show Hint
Find the side lengths opposite each vertex and use the incentre formula with weights equal to those lengths.
Updated On: Nov 12, 2026
- \(2\hat{i}+3\hat{k}\)
- \(2\hat{i}-3\hat{k}\)
- \(2\hat{j}-3\hat{k}\)
- \(2\hat{j}+3\hat{k}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Side lengths
\(a=BC=3\), \(b=CA=4\), \(c=AB=5\).
Step 2: Incentre formula
\(I=\frac{aA+bB+cC}{a+b+c}\).
Step 3: Substitute
\(I=\frac{3(0,3,0)+4(0,0,4)+5(0,3,4)}{12}=\frac{(0,24,36)}{12}\).
Step 4: Result
\(I=(0,2,3)\), that is \(2\hat{j}+3\hat{k}\).
Final Answer:
The incentre is \(2\hat{j}+3\hat{k}\). \[ \boxed{\text{(D) }2\hat{j}+3\hat{k}} \]
\(a=BC=3\), \(b=CA=4\), \(c=AB=5\).
Step 2: Incentre formula
\(I=\frac{aA+bB+cC}{a+b+c}\).
Step 3: Substitute
\(I=\frac{3(0,3,0)+4(0,0,4)+5(0,3,4)}{12}=\frac{(0,24,36)}{12}\).
Step 4: Result
\(I=(0,2,3)\), that is \(2\hat{j}+3\hat{k}\).
Final Answer:
The incentre is \(2\hat{j}+3\hat{k}\). \[ \boxed{\text{(D) }2\hat{j}+3\hat{k}} \]
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