This question is similar Chapter 4 Class 12 Determinants - Ex 4.2

Please check the question here 

https://www.teachoo.com/3229/690/Ex-4.3--2---Show-that-A-(a---b---c)--B-(b-c---a)--C-(c-a---b)/category/Ex-4.3/

 

Slide12.JPG

Slide13.JPG
Slide14.JPG

 

Go Ad-free

Transcript

Question 6 If the points (π‘₯_1,𝑦_1 ),(π‘₯_2,𝑦_2 ) and (π‘₯_1+π‘₯_2,𝑦_1+𝑦_2 ) are collinear, then π‘₯_1 𝑦_2 is equal to (A) π‘₯_2 𝑦_1 (B) π‘₯_1 𝑦_1 (C) π‘₯_2 𝑦_2 (D) π‘₯_1 π‘₯_2Three point are collinear if they lie on some line 𝑖.𝑒. They do not form a triangle ∴ Area of triangle = 0 We know that Area of triangle is given by βˆ† = 1/2 |β– 8(x1&y1&1@x2&y2&1@x3&y3&1)| Here, x1 = x1, y1 = y1 x2 = x2, y2 = y2, x3 = x1 + x2, y3 = y1 + y2 Putting values βˆ† = 1/2 |β– 8(π‘₯_1&𝑦_1&1@π‘₯_2&𝑦_2&1@π‘₯_1+π‘₯_2&𝑦_1+𝑦_2&1)| βˆ† = 1/2[π‘₯_1 (𝑦_2 Γ— 1βˆ’(𝑦_1+𝑦_2 )Γ— 1) βˆ’ 𝑦_1 (π‘₯_2 Γ—1 βˆ’(π‘₯_1+π‘₯_2 )Γ—1) +1(π‘₯_2 Γ—(𝑦_1+𝑦_2 )βˆ’(π‘₯_1+π‘₯_2 )×𝑦_2 ) ] βˆ† = 1/2[π‘₯_1 (𝑦_2 βˆ’π‘¦_1βˆ’π‘¦_2 ) βˆ’ 𝑦_1 (π‘₯_2 βˆ’π‘₯_1βˆ’π‘₯_2 ) + 1(π‘₯_2 𝑦_1+π‘₯_2 𝑦_2βˆ’π‘₯_1 𝑦_2βˆ’π‘₯_2 𝑦_2 ) ] βˆ† = 1/2[βˆ’π‘₯_1 𝑦_1+π‘₯_1 𝑦_1+π‘₯_2 𝑦_1βˆ’π‘₯_1 𝑦_2] βˆ† = 1/2[π‘₯_2 𝑦_1βˆ’π‘₯_1 𝑦_2] Putting Area of Triangle = βˆ† = 0 0 = 1/2[π‘₯_2 𝑦_1βˆ’π‘₯_1 𝑦_2] 0 = π‘₯_2 𝑦_1βˆ’π‘₯_1 𝑦_2 π‘₯_1 𝑦_2=𝒙_𝟐 π’š_𝟏 So, the correct answer is (A)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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 and Computer Science at Teachoo