Class CBSE Class 12 Mathematics Matrices and Determinants Q #1688
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
If the area of $\Delta$ ABC with vertices $A(3,1)$, $B(-2,1)$ and $C(0,k)$ is 5 sq. units, then values of k are:
(A) 3, 1
(B) -1, 3
(C) -1, 2
(D) 0, 2
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Correct Answer: B

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Detailed Solution

Step 1: Recall the Area Formula

The area of a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ is given by the determinant formula: $$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Step 2: Substitute the given values

Given vertices are $A(3, 1)$, $B(-2, 1)$, and $C(0, k)$. The area is 5 sq. units. $$5 = \frac{1}{2} |3(1 - k) + (-2)(k - 1) + 0(1 - 1)|$$

Step 3: Simplify the expression

Multiply by 2 and simplify the terms inside the modulus: $$10 = |3 - 3k - 2k + 2|$$ $$10 = |5 - 5k|$$

Step 4: Solve for k

This implies two cases: Case 1: $5 - 5k = 10 \implies -5k = 5 \implies k = -1$ Case 2: $5 - 5k = -10 \implies -5k = -15 \implies k = 3$

Final Answer: -1, 3

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must utilize the determinant-based area formula to solve for an unknown coordinate variable.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the execution of a specific mathematical algorithm (determinant expansion) to find the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This aligns with the Determinants chapter, testing the ability to apply properties of determinants to geometric problems.
||KEY:B

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