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Question

Mathematics Question on Determinants

Find the area of the triangle with vertices P(4,5)P(4, 5), Q(4,2)Q(4, -2) and R(6,2)R(-6, 2).

A

21 s units

B

35 s units

C

30 s units

D

40 s units

Answer

35 s units

Explanation

Solution

Let Δ\Delta be the area of triangle PQRPQR. Then, Δ=12451 421 621\Delta=\frac{1}{2}\left|\begin{matrix}4&5&1\\\ 4&-2&1\\\ -6&2&1\end{matrix}\right| Applying R1R1R2,R2R2R3R_{1} \rightarrow R_{1}-R_{2}, R_{2}\rightarrow R_{2}-R_{3}, we get Δ=12070 1040 621\Delta=\frac{1}{2}\left|\begin{matrix}0&7&0\\\ 10&-4&0\\\ -6&2&1\end{matrix}\right| Expanding along R1R_{1}, we get Δ=12[7(100)]=1270=35\Rightarrow\quad\Delta=\frac{1}{2}\left|\left[-7\left(10-0\right)\right]\right|=\frac{1}{2}\left|-70\right|=35 s units