Question

Let \[ \det A=\left|\begin{matrix}l&m&n\\ p&q&r\\ 1&1&1\end{matrix}\right|. \] If \[ (l-m)^{2}+(p-q)^{2}=9, \] \[ (m-n)^{2}+(q-r)^{2}=16, \] \[ (n-l)^{2}+(r-p)^{2}=25, \] then the value of \((\det A)^{2}\) is:

• A. 169
• B. 144
• C. 121
• D. 100

Hint:

Whenever a coordinate geometry problem features the numbers 9, 16, and 25, it is a strong hint that you are working with a classic 3-4-5 right-angled triangle. Recognizing this geometric shape immediately saves you from doing long algebraic expansions.

Answer 1

The Correct Option is B

Concept: This problem can be solved by interpreting the matrix parameters geometrically. In coordinate geometry, a determinant of this specific format represents exactly twice the area of a triangle whose vertices are located at the coordinate points $P_1(l, p)$, $P_2(m, q)$, and $P_3(n, r)$ in a 2D plane: $$\text{Area} = \frac{1}{2} \cdot \det A \quad \Rightarrow \quad (\det A)^2 = 4 \cdot (\text{Area})^2$$ Step 1: Interpret the equations as distance constraints.
Let us analyze the three given squared equations using the standard Euclidean distance formula between two points, $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$:
• Distance squared between $P_1(l,p)$ and $P_2(m,q)$: $a^2 = (l-m)^2 + (p-q)^2 = 9 \implies a = 3$
• Distance squared between $P_2(m,q)$ and $P_3(n,r)$: $b^2 = (m-n)^2 + (q-r)^2 = 16 \implies b = 4$
• Distance squared between $P_3(n,r)$ and $P_1(l,p)$: $c^2 = (n-l)^2 + (r-p)^2 = 25 \implies c = 5$

Step 2: Identify the geometric shape of the triangle.
The side lengths of our triangle are $a = 3$, $b = 4$, and $c = 5$. Notice that these side lengths satisfy the Pythagorean theorem: $$a^2 + b^2 = 3^2 + 4^2 = 9 + 16 = 25 = c^2$$ This confirms that the vertices form a right-angled triangle where the sides of length 3 and 4 meet at a $90^\circ$ angle, and the side of length 5 is the hypotenuse.

Step 3: Calculate the geometric area of the triangle.
Using the standard area formula for a right-angled triangle ($\frac{1}{2} \times \text{base} \times \text{height}$): $$\text{Area} = \frac{1}{2} \times 3 \times 4 = 6$$

Step 4: Calculate the value of $(\det A)^2$.
Using our core geometric relationship from the concept section: $$(\det A)^2 = 4 \cdot (\text{Area})^2 = 4 \cdot (6)^2 = 4 \times 36 = 144$$ This matches option (B) perfectly.