# Ex 7.3, 2

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 7.3 , 2 In each of the following find the value of ‘k’, for which the points are collinear. (7, –2), (5, 1), (3, k) Let the given points be A(7, −2) , B(5, 1) , C(3, k) If the above points are collinear, they will lie on the same line, i.e. the will not form triangle or We can say that Area of ∆ABC = 0 ⇒ 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] = 0 Here x1 = 7 , y1 = −2 x2 = 5 , y2 = 1 x3 = 3 , y3 = k Putting values 1/2 [ 7(1 − k) + 5(k – (−2)) + 3(−2 − 1) ] = 0 7(1 − k) + 5(k +2) + 3(−3) = 0 × 2 7 – 7k + 5k + 10 – 9 = 0 – 7k + 5k = − 10 + 9 – 7 −2k = −8 −2k = −8 k = (−8)/(−2) k = 4 Ex 7.3 , 2 In each of the following find the value of ‘k’, for which the points are collinear. (ii) (8, 1), (k, – 4), (2, –5) Let the given points be A(8, 1) , B(k, −4) , C(2, −5) If the above points are collinear, they will lie on the same line, i.e. the will not form triangle or We can say that Area of ∆ABC = 0 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] = 0 Here x1 = 8 , y1 = 1 x2 = k , y2 = −4 x3 = 2 , y3 = −5 Putting values 1/2 [ 8(−4 – (−5)) + k(−5 − 1) + 2(1 – (−4)) ] = 0 8(−4 + 5) + k(−6) + 2(1 + 4) = 0 × 2 8(1) + k(−6) + 2(5) = 0 8 −6k + 10 = 0 −6k = −10 – 8 −6k = −10 − 8 −6k = −18 k = (−18)/(−6) k = 3

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .