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

Open in App

Solution

$∆ = \frac{1}{2} |\begin{matrix} 1 & 4 & 1 \\ 2 & 3 & 1 \\ - 5 & - 3 & 1 \end{matrix}| = \frac{1}{2} |\begin{matrix} 1 & 4 & 1 \\ 1 & - 1 & 0 \\ - 5 & - 3 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{1}] = \frac{1}{2} |\begin{matrix} 1 & 4 & 1 \\ 1 & - 1 & 0 \\ - 6 & - 7 & 0 \end{matrix}| [Applying R_{3} \to R_{3} - R_{1}] = \frac{1}{2} |\begin{matrix} 1 & - 1 \\ - 6 & - 7 \end{matrix}| = \frac{1}{2} (- 7 - 6) = \frac{13}{2} square units [∵ Area cannot be negative]$

Therefore, (1, 4), (2, 3) and (−5, −3) are not collinear because, $|\begin{matrix} 1 & 4 & 1 \\ 2 & 3 & 1 \\ - 5 & - 3 & 1 \end{matrix}|$ is not equal to 0.

Suggest Corrections

Similar questions

Assertion: Area of the triangle whose
vertices are $A (\frac{- 3}{2}, 3), B (6, - 2)$ and $C (- 3, 4)$ is 0
Reason : The points A, B and C are collinear

Q. Find the area of region bounded by the triangle whose vertices are

(- 1, 1), (0, 5)

and

(3, 2)

, using integration.

Q. Given three vertices of a triangle whose coordinates are A (1, 1), B (3, -3) and (5, -3). Find the area of the triangle.

Q. Using determinent, if area of triangle is 4, whose vertices are (2, 2), (6, 6), (5, k) then k=______.

Join BYJU'S Learning Program

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

∆=12141231-5-31=121 4 11-10-5-31 Applying R2→R2-R1=121411-10-6-70 Applying R3→R3-R1=121-1-6-7=12-7 - 6=132square units ∵Area cannot be negative Therefore, (1, 4), (2, 3) and (−5, −3) are not collinear because, 141231-5-31 is not equal to 0.