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
CBSE Class 12 Sample Paper for 2024 Boards
CBSE Class 12 Sample Paper for 2024 Boards
Last updated at August 14, 2026 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)