If the area of \( \triangle ABC \) with vertices \( A(3, 1) \), \( B(-2, 1) \) and \( C(0, k) \) is 5 sq. units, then values of \( k \) are :
Show Hint
- \( 3, 1 \)
- \( -1, 3 \)
- \( -1, 2 \)
- \( 0, 2 \)
The Correct Option is B
Solution and Explanation
Concept:
The area of a triangle with vertices \( (x_1,y_1) \), \( (x_2,y_2) \), and \( (x_3,y_3) \) is given by: \[ \text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| \]
Step 1: Substitute the given coordinates
Given \(A(3,1)\), \(B(-2,1)\), \(C(0,k)\), and area \(=5\): \[ 5=\frac{1}{2}\left|3(1-k)+(-2)(k-1)+0(1-1)\right| \] \[ 10=\left|3-3k-2k+2\right| \]
Step 2: Simplify the equation
\[ 10=|5-5k| \] Dividing both sides by \(5\): \[ 2=|1-k| \]
Step 3: Solve the absolute value equation
For \( |1-k|=2 \), we have two cases:
Case 1: \[ 1-k=2 \] \[ k=-1 \] Case 2: \[ 1-k=-2 \] \[ k=3 \]
Final Answer:
Therefore, the possible values of \(k\) are: \[ \boxed{k=-1,\ 3} \]
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