If area of the triangle formed by A(1, 3), B(0, 0) and C(k, 0) is 3 sq. units, then which of the following is/are TRUE?
A. k = 2
B. k = −2
C. k = 3
D. k = −3
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A and B only
By the determinant formula for the area of a triangle:
\( \displaystyle \Delta=\frac{1}{2}\begin{vmatrix}1 & 3 & 1\\0 & 0 & 1\\k & 0 & 1\end{vmatrix} \)
Expanding along the second row:
\( \displaystyle \begin{vmatrix}1 & 3 & 1\\0 & 0 & 1\\k & 0 & 1\end{vmatrix}=-1\begin{vmatrix}1 & 3\\k & 0\end{vmatrix} \)
\( \displaystyle =-(0-3k)=3k \)
Area is the absolute value, so
\( \displaystyle \frac{1}{2}|3k|=3 \)
\( \displaystyle |k|=2 \)
So k = 2 or k = −2.
Checking the statements
(A) k = 2 — Correct. \( \frac{1}{2}\times 2\times 3=3 \).
(B) k = −2 — Correct. \( \frac{1}{2}\times|-2|\times 3=3 \).
(C) k = 3 — Incorrect. The area would be \( \frac{1}{2}\times 3\times 3=4.5 \).
(D) k = −3 — Incorrect. The area would be \( \frac{1}{2}\times|-3|\times 3=4.5 \).
Shortcut: B and C lie on the x-axis, so BC = |k| is the base and the height is the y-coordinate of A, which is 3. Then \( \frac{1}{2}\times|k|\times 3=3 \) gives |k| = 2.
NCERT (2026-27 edition):
Area of a Triangle
