For every real number k, the area of the triangle with vertices \((k, k^{2})\), \((k + 1, (k + 1)^{2})\) and \((k + 3, (k + 3)^{2})\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3 square units
The area is the absolute value of
\(\displaystyle \Delta = \frac{1}{2}\begin{vmatrix}k & k^{2} & 1\\ k + 1 & (k + 1)^{2} & 1\\ k + 3 & (k + 3)^{2} & 1\end{vmatrix}\)
The row operations \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\) give
\((k + 1)^{2} - k^{2} = 2k + 1\)
\((k + 3)^{2} - k^{2} = 6k + 9\)
\(\displaystyle \Delta = \frac{1}{2}\begin{vmatrix}k & k^{2} & 1\\ 1 & 2k + 1 & 0\\ 3 & 6k + 9 & 0\end{vmatrix}\)
Expanding along the third column, only the first entry is non-zero:
\(\displaystyle \Delta = \frac{1}{2}\left[1(6k + 9) - 3(2k + 1)\right]\)
\(\displaystyle \Delta = \frac{1}{2}(6k + 9 - 6k - 3) = \frac{1}{2}(6) = 3\)
The k-terms cancel, so the area is 3 square units for every value of k.
Check with k = 0: the vertices (0, 0), (1, 1), (3, 9) give \(\frac{1}{2}\left|0 + 1(9 - 0) + 3(0 - 1)\right| = 3\).
Why the other options are incorrect
Option (1) 6 square units — Incorrect. 6 is the value of the determinant; the area is half of it.
Option (2) \((k^{2} + 3)\) square units — Incorrect. The k-terms cancel; for k = 1 the vertices (1, 1), (2, 4), (4, 16) still give area 3, not 4.
Option (4) \(\displaystyle\frac{3}{2}\) square units — Incorrect. This halves the area 3 a second time.