Let A (1, 3), B (0, 0) and C (k, 0) be three points such that area of \( \Delta(ABC) \) = 3 sq. units, then the value of 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
\( \displaystyle \Delta=\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix} \)
For A(1, 3), B(0, 0) and C(k, 0):
\( \displaystyle \Delta=\frac{1}{2}\begin{vmatrix}1 & 3 & 1\\0 & 0 & 1\\k & 0 & 1\end{vmatrix} \)
\( \displaystyle =\frac{1}{2}\left[1(0-0)-3(0-k)+1(0-0)\right] \)
\( \displaystyle =\frac{3k}{2} \)
Since the area is 3, the determinant value may be 3 or −3:
\( \displaystyle \left|\frac{3k}{2}\right|=3 \)
\( |k|=2 \)
\( k=\pm 2 \)
Shortcut: B and C lie on the x-axis, so take BC as the base: base = |k| and height = 3 (the y-coordinate of A).
\( \displaystyle \frac{1}{2}\times|k|\times 3=3\ \Rightarrow\ |k|=2 \)
NCERT (2026-27 edition):
Area of a Triangle
