CBSE Class 12 Sample Paper for 2024 Boards
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CBSE Class 12 Sample Paper for 2024 Boards
Last updated at Feb. 10, 2025 by Teachoo
This question is similar to Ex 4.2, 3(i) Chapter 4 Class 12
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)
