if the area of the triangle formed by the points (2a,b)(a+b,2b+a) and (2b,2a) be ? then the area of the triangle whose vertices are (a+b,a-b) , (3b-a,b+3a) and (3a-b,3b-a) is

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Solution

Dear student
$W e k n o w t h a t T h e a r e a o f t h e t r i a n g l e, t h e c o o r d i n a t e s o f w h o s e v e r t i c e s a r e (x_{1}, y_{1}), (x_{2}, y_{2}) a n d (x_{3}, y_{3}) i s \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| N o w i f c o o r d i n a t e s a r e (2 a, b), (a + b, 2 b + a) a n d (2 b, 2 a) t h e n a r e a i s λ = \frac{1}{2} |2 a (2 b + a - 2 a) + (a + b) (2 a - b) + 2 b (b - 2 b - a)| [G i v e n] \Rightarrow λ = \frac{1}{2} |2 a (2 b - a) + (a + b) (2 a - b) + 2 b (- b - a)| \Rightarrow λ = \frac{1}{2} |4 a b - 2 a^{2} + 2 a^{2} - a b + 2 a b - b^{2} - 2 b^{2} - 2 a b| \Rightarrow λ = \frac{1}{2} |3 a b - 3 b^{2}| . . . (1) N o w i f c o o r d i n a t e s a r e (a + b, a - b), (3 b - a, b + 3 a) a n d (3 a - b, 3 b - a) t h e n a r e a i s △ = \frac{1}{2} |(a + b) (b + 3 a - 3 b + a) + (3 b - a) (3 b - a - a + b) + (3 a - b) (a - b - b - 3 a)| = \frac{1}{2} |(a + b) (- 2 b + 4 a) + (3 b - a) (4 b - 2 a) + (3 a - b) (- 2 b - 2 a)| = \frac{1}{2} |- 2 a b + 4 a^{2} - 2 b^{2} + 4 a b + 12 b^{2} - 6 a b - 4 a b + 2 a^{2} - 6 a b - 6 a^{2} + 2 b^{2} + 2 a b| = \frac{1}{2} |- 12 a b + 12 b^{2}| = \frac{1}{2} |- 4 (3 a b - 3 b^{2})| = 4 \times \frac{1}{2} |3 a b - 3 b^{2}| = 4 λ [f r o m e q (1)]$
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