Question

If \(A(x_1,y_1)\), \(B(x_2,y_2)\) and \(C(x_3,y_3)\) are the vertices of an equilateral triangle whose each side is equal to \(k\), then the value of \(\begin{vmatrix} x_1 & y_1 & 2 \\ x_2 & y_2 & 2 \\ x_3 & y_3 & 2 \end{vmatrix}^2\) is

• A. \(3k^2\)
• B. \(4k^2\)
• C. \(3k^4\)
• D. \(4k^4\)

Hint:

Take 2 common from the last column. The determinant with 1s gives twice the area.

Answer 1

The Correct Option is C

Step 1: Recall the area formula.
The area of a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\) is half the absolute value of the determinant with a column of 1s.
\[ \Delta = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right| \]

Step 2: Pull out the common factor.
Every entry of the last column is 2. Taking 2 out of that column gives a factor of 2 outside the determinant.
\[ \begin{vmatrix} x_1 & y_1 & 2 \\ x_2 & y_2 & 2 \\ x_3 & y_3 & 2 \end{vmatrix} = 2\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]

Step 3: Use the area of an equilateral triangle.
For side \(k\), the area is \(\Delta = \frac{\sqrt{3}}{4}k^2\). From Step 1, the determinant with 1s equals \(\pm 2\Delta = \pm\frac{\sqrt{3}}{2}k^2\).

Step 4: Find the required value.
The given determinant is \(2 \times \left(\pm\frac{\sqrt{3}}{2}k^2\right) = \pm\sqrt{3}k^2\). Squaring removes the sign.
\[ \left(\sqrt{3}k^2\right)^2 = 3k^4 \]

Step 5: Check the options.
Option 1 and option 2 have only \(k^2\), which is wrong because squaring a term with \(k^2\) must give \(k^4\). Option 4 would need the determinant to be \(2k^2\), but it is \(\sqrt{3}k^2\). So only option 3 fits.

Final Answer:
The value is \(3k^4\), which is option 3. \[ \boxed{3k^4} \]