Ex 4.3,1 (i) - Chapter 4 Class 12 Determinants (Term 1)
Last updated at Dec. 23, 2021 by Teachoo
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Ex 4.3,1
Find area of the triangle with vertices at the point given in each of the following:
(1, 0), (6, 0), (4, 3)
The area of triangle is given by ∆ = 𝟏/𝟐 |■8(𝐱𝟏&𝐲𝟏&𝟏@𝐱𝟐&𝐲𝟐&𝟏@𝐱𝟑&𝐲𝟑&𝟏)|
Here,
x1 = 1 , y1 = 0
x2 = 6 ,y2 = 0
x3 = 4 ,y3 = 3
∆ = 𝟏/𝟐 |■8(𝟏&𝟎&𝟏@𝟔&𝟎&𝟏@𝟒&𝟑&𝟏)|
= 1/2 (1|■8(0&1@3&1)|−0|■8(6&1@4&1)|+1|■8(6&0@4&3)|)
= 1/2 (1(0 – 3) – 0(6 – 4) + 1 (18 – 0))
= 1/2 (1(–3) + 0 + 1 (18) )
= 1/2 [–3 + 18 ]
= 𝟏𝟓/𝟐
Thus, the required area of triangle is 𝟏𝟓/𝟐 square units
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Davneet Singh
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