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: Use the centroid formula.
For a triangle with vertices \[ A(x_1,y_1,z_1),\quad B(x_2,y_2,z_2),\quad C(x_3,y_3,z_3), \] the centroid is \[ \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}\right) \] Here, \[ A(1,2,3),\quad B(h,-3,0),\quad C(-4,k,-1) \] Given centroid is \[ \left(5,-1,\frac{2}{3}\right) \]

Step 2: Find the value of \(h\).
Using the \(x\)-coordinate of centroid, \[ \frac{1+h-4}{3}=5 \] \[ \frac{h-3}{3}=5 \] \[ h-3=15 \] \[ h=18 \]

Step 3: Find the value of \(k\).
Using the \(y\)-coordinate of centroid, \[ \frac{2-3+k}{3}=-1 \] \[ \frac{k-1}{3}=-1 \] \[ k-1=-3 \] \[ k=-2 \] Thus, the vertices are \[ A(1,2,3),\quad B(18,-3,0),\quad C(-4,-2,-1) \]

Step 4: Find the squares of the side lengths.
Now, \[ AB^2=(18-1)^2+(-3-2)^2+(0-3)^2 \] \[ AB^2=17^2+(-5)^2+(-3)^2 \] \[ AB^2=289+25+9=323 \] Similarly, \[ AC^2=(-4-1)^2+(-2-2)^2+(-1-3)^2 \] \[ AC^2=(-5)^2+(-4)^2+(-4)^2 \] \[ AC^2=25+16+16=57 \] Also, \[ BC^2=(-4-18)^2+(-2+3)^2+(-1-0)^2 \] \[ BC^2=(-22)^2+1^2+(-1)^2 \] \[ BC^2=484+1+1=486 \]

Step 5: Check the type of triangle.
The greatest side square is \[ BC^2=486 \] Now compare it with the sum of the other two squares: \[ AB^2+AC^2=323+57=380 \] Since \[ BC^2\gt AB^2+AC^2 \] \[ 486\gt 380 \] Therefore, the triangle is an obtuse angled triangle.

Step 6: Final conclusion.
Hence, triangle \(ABC\) is \[ \boxed{\text{an obtuse angled triangle}} \]