Question 16 (OR 2 nd  question)

Find the value of k for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, –k – 2) are collinear.   Subscribe to our Youtube Channel - https://you.tube/teachoo

1. Class 10
2. Solutions of Sample Papers for Class 10 Boards
3. CBSE Class 10 Sample Paper for 2019 Boards

Transcript

Question 16 (OR 2nd question) Find the value of k for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, –k – 2) are collinear. Let points be A (3k – 1, k – 2), B (k, k – 7) and C (k – 1, –k – 2) If A, B, C are collinear, they will lie on the same line, i.e. they will not form triangle Therefore, Area of ∆ABC = 0 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] = 0 1 mark Here x1 = 3k – 1 , y1 = k – 2 x2 = k , y2 = k – 7 x3 = k – 1 , y3 = −k – 2 Putting values 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] = 0 x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0 (3k – 1)[k – 7 – (–k – 2)] + k[−k − 2 – (k – 2)] + (k – 1)[(k – 2) – (k – 7)] = 0 (3k – 1)[k – 7 + k + 2] + k[−k − 2 – k + 2] + (k – 1)[k – 2 – k + 7] = 0 (3k – 1)(2k – 5) + k[−2k] + (k – 1)[–2 + 7] = 0 3k(2k – 5) – 1(2k – 5) – 2k2 + (k – 1)5 = 0 6k2 – 15k – 2k + 5 – 2k2 + 5k – 5 = 0 6k2 – 2k2 – 15k – 2k + 5k + 5 – 5 = 0 4k2 – 12k = 0 4k(k – 3) = 0 1 mark So, k = 0 and k = 3 1 mark

CBSE Class 10 Sample Paper for 2019 Boards

Class 10
Solutions of Sample Papers for Class 10 Boards 