If the area of a triangle with vertices A(1,3), B(0,0) and C(k,0) is 3 sq. units, then k is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\pm 2\)
The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is the absolute value of
\( \displaystyle \Delta = \frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix} \)
With A(1, 3), B(0, 0), C(k, 0):
\( \displaystyle \Delta = \frac{1}{2}\begin{vmatrix}1 & 3 & 1\\ 0 & 0 & 1\\ k & 0 & 1\end{vmatrix} \)
Expanding along the second row (only its third entry is non-zero):
\( \displaystyle \Delta = \frac{1}{2}\left(-1\right)\begin{vmatrix}1 & 3\\ k & 0\end{vmatrix} \)
\( \displaystyle = \frac{1}{2}\left(-1\right)(0 - 3k) = \frac{3k}{2} \)
The area is 3, so
\( \displaystyle \frac{3k}{2} = \pm 3 \)
\(k = \pm 2\)
Both signs are needed because the determinant gives the area with a sign.
NCERT (2026-27 edition):
Area of a Triangle
