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Question #40755

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(x_A, y_A)$ , $B(x_B, y_B)$ , and $C(x_C, y_C)$ be the vertices of a triangle given in counter-clockwise order, with an area of $18$ sq. units. Point $P$ lies on the line segment $BC$ such that $BP:PC = 1:2$ . Point $Q$ lies on the line $CA$ outside the segment $CA$ such that $CQ:QA = 2:1$ . Point $R$ lies on the line segment $AB$ such that $AR:RB = 1:2$ . Find the value of the determinant $\Delta = \begin{vmatrix} x_P & x_Q & x_R \\ y_P & y_Q & y_R \\ 1 & 1 & 1 \end{vmatrix}$ .

Options

A.
$-8$
B.
$8$
C.
$-4$
D.
$4$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Internal vs External Division|Determinant as Geometric Area/Volume|Row/Column Exchange Parity

Why This Question Is Tricky

Common Mistakes

This question tests the understanding of determinants as directed areas and the effect of affine transformations on geometric orientation.

A common trap is calculating the absolute area of the new triangle (which is 4) and assuming the determinant is simply $2 \times \text{Area} = 8$ .

However, the external division causes the orientation of $\triangle PQR$ to be clockwise (reversed from the counter-clockwise $\triangle ABC$ ).

This orientation reversal is mathematically captured by the negative determinant of the transformation matrix $T$ .

Failing to account for this parity change leads to the incorrect positive sign.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Internal vs External Division
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free