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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Solution

$A r e a = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 4 & 1 2 & 3 & 1 - 5 & - 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ A r e a = \frac{1}{2} | {1 (3 + 3) - 4 (2 + 5) + 1 (- 6 + 15)} | A r e a = \frac{15}{2}$
No given points are not collinear.

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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. Using integration, find the area of the triangle whose vertices are

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and

(4, 4) .

Q. Using vectors, find the area of the triangle ABC, whose vertices are

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and

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(ii) (−2, 3) (2, −1) (4, 0)

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

Area=12∣∣ ∣∣141231−5−31∣∣ ∣∣Area=12|{1(3+3)−4(2+5)+1(−6+15)}|Area=152No given points are not collinear.

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

$A r e a = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 4 & 1 2 & 3 & 1 - 5 & - 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ A r e a = \frac{1}{2} | {1 (3 + 3) - 4 (2 + 5) + 1 (- 6 + 15)} | A r e a = \frac{15}{2}$
No given points are not collinear.