Let A be the area of the triangle with vertices (3, 8), (5, -1) and \( (k,2),k\in\mathrm{Z} \) (set of integers). If \( 2A=75 \), then \( k \) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → -4
Twice the area of a triangle is the absolute value of the determinant of its vertices:
\( \displaystyle 2A=\left|\begin{vmatrix}3 & 8 & 1\\5 & -1 & 1\\k & 2 & 1\end{vmatrix}\right| \)
\( \displaystyle \begin{vmatrix}3 & 8 & 1\\5 & -1 & 1\\k & 2 & 1\end{vmatrix}=3(-1-2)-8(5-k)+1(10+k) \)
\( \displaystyle =-9-40+8k+10+k \)
\( \displaystyle =9k-39 \)
\( \displaystyle |9k-39|=75\ \Rightarrow\ 9k-39=75\ \text{or}\ 9k-39=-75 \)
\( \displaystyle 9k=114\ \Rightarrow\ k=\frac{38}{3}\ \text{(not an integer)} \)
\( \displaystyle 9k=-36\ \Rightarrow\ k=-4 \)
Since \( k\in\mathrm{Z} \), k = −4.
Shortcut: Put the options into \( 9k-39 \): k = −4 gives −75, k = 4 gives −3, k = 13 gives 78, and k = −21 gives −228. Only k = −4 gives a modulus of 75.