मराठी

Using Determinants, Find the Area of the Triangle Whose Vertices Are (1, 4), (2, 3) and (−5, −3). Are the Given Points Collinear?

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प्रश्न

Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?

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उत्तर

\[∆ = \frac{1}{2}\begin{vmatrix}1 & 4 & 1 \\ 2 & 3 & 1 \\ - 5 & - 3 & 1\end{vmatrix}\] 
\[ = \frac{1}{2}\begin{vmatrix}1 & 4 & 1 \\ 1 & - 1 & 0 \\ - 5 & - 3 & 1\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\]
\[ = \frac{1}{2}\begin{vmatrix}1 & 4 & 1 \\ 1 & - 1 & 0 \\ - 6 & - 7 & 0\end{vmatrix} \left[\text{ Applying }R_3 \to R_3 - R_1 \right]\] 
\[ = \frac{1}{2}\begin{vmatrix}1 & - 1 \\ - 6 & - 7\end{vmatrix}\]
\[ = \frac{1}{2}\left( - 7 - 6 \right)\] 
\[ = \frac{13}{2}\text{ square units }\left[ \because\text{ Area cannot be negative }\right]\]
Therefore, (1, 4), (2, 3) and (−5, −3) are not collinear because,

\[\begin{vmatrix}1 & 4 & 1 \\ 2 & 3 & 1 \\ - 5 & - 3 & 1\end{vmatrix}\]  is not equal to 0.
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पाठ 5: Determinants - Exercise 6.3 [पृष्ठ ७१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 5 Determinants
Exercise 6.3 | Q 7 | पृष्ठ ७१

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