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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 4.2, 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.