Question

If the vertices of a triangle \(ABC\) are \(A(1,2,3)\), \(B(h,-3,0)\) and \(C(-4,k,-1)\) and the centroid of the triangle is \(\left(5,-1,\dfrac{2}{3}\right)\), then triangle \(ABC\) is:

• A. an obtuse angled triangle
• B. an acute angled triangle
• C. an isosceles triangle
• D. a right angled triangle

Hint:

For classifying a triangle using side lengths: \[ \text{If largest side}^2\gt \text{sum of squares of other two sides}, \] then the triangle is obtuse angled.

Answer 1

The Correct Option is A

Step 1: Find h and k using the centroid formula.
The centroid of triangle ABC with $ A(1,2,3) $, $ B(h,-3,0) $, $ C(-4,k,-1) $ equals $ (5,-1,2/3) $. So: \[ \frac{1+h-4}{3}=5 \implies h=18, \quad \frac{2-3+k}{3}=-1 \implies k=-2 \]
Step 2: Compute the side vectors from A.
\[ \vec{AB} = (17,-5,-3), \quad \vec{AC} = (-5,-4,-4) \]
Step 3: Compute the dot product to find the angle at A.
\[ \vec{AB}\cdot\vec{AC} = (17)(-5)+(-5)(-4)+(-3)(-4) = -85+20+12 = -53 \] Since the dot product is negative, angle A is obtuse.
Step 4: Compute all three squared side lengths.
\[ AB^2 = 289+25+9 = 323, \quad AC^2 = 25+16+16 = 57, \quad BC^2 = 484+1+1 = 486 \]
Step 5: Confirm using the cosine rule.
\[ \cos A = \frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC} = \frac{380-486}{2\sqrt{323\times 57}} < 0 \] This confirms angle $ A > 90^\circ $.
Step 6: Classify the triangle.
Since one angle is obtuse, this is an obtuse-angled triangle. \[ \boxed{\text{Obtuse angled triangle}} \]