If the area of the triangle with vertices (-3,0),(3,0) and (0,k) is 9 sq units, then the value/s of k will be

(a) 9                               (b) ± 3                             (c) -9                         (d) 6

This question is similar to Ex 4.2, 3(i) Chapter 4 Class 12

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Transcript

https://www.teachoo.com/3230/690/Ex-4.3--3---Find-values-of-k-if-area-of-triangle-is-4--vertices/category/Ex-4.3/x1 = −3 , y1 = 0, x2 = 3, y2 = 0, x3 = 0 y3 = k Our equation becomes ±9 = 𝟏/𝟐 |■8(−𝟑&𝟎&𝟏@𝟑&𝟎&𝟏@𝟎&𝐤&𝟏)| ± 9 = 1/2 (−3|■8(0&1@𝑘&1)|−0|■8(3&1@0&1)|+1|■8(3&0@0&𝑘)|) ± 9 = 1/2 [−3 (0 − k) – 0(3 – 0) + 1 (3k – 0)] ± 9 × 2 = (−3 (–k) – 0 + 1 (3k)) ± 18 = 3k + 3k ± 18 = 6k k = ±𝟑 So, the correct answer is (b)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.