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Ex 10.1, 9 - Without using distance formula, show that - Ex 10.1

Ex 10.1, 9 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.1, 9 - Chapter 10 Class 11 Straight Lines - Part 3 Ex 10.1, 9 - Chapter 10 Class 11 Straight Lines - Part 4


Transcript

Ex 10.1, 9 Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and (–3, 2) are vertices of a parallelogram. Let the given points be A (–2, –1) , B (4, 0), C (3, 3) , D (–3, 2) We have to prove if ABCD is a parallelogram ABCD is a parallelogram if both pairs of opposite sides are parallel i.e. AB ∥ CD & AD ∥ BC So, we have to prove Slope of AB = Slope of CD Slope of AD = Slope of BC Lets calculate slope of AB, BC, CD, and AD Slope of AB A (–2, –1), B (4, 0) Here, x1 = –2, y1 = –1 x2 = 4, y2 = 0 Slope of AB = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) = (0 − (−1))/(4 − (−2)) = 1/(4 + 2) = 1/6 Slope of CD C (3, 3), D (–3, 2) Here, x1 = 3, y1 = 3 x2 = –3, y2 = 2 Slope of BC = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) = (2 − 3)/(−3 − 3) = (−1)/( −6) = 1/6 Since Slope of AB = Slope of CD ∴ AB ∥ CD Slope of AD A (–2, –1) , B (–3, 2) Here, x1 = –2, y1 = –1 x2 = –3, y2 = 2 Slope of AB = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) = (2 − (−1))/(−3 − (−2)) = (2 + 1)/(−3 + 2) = 3/(−1) = –3 Slope of BC B (4, 0) , C (3, 3) Here, x1 = 4, y1 = 0 x2 = 3, y2 = 3 Slope of BC = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) = (3 − 0)/(3 − 4) = 3/(−1) = −3 Since Slope of AD = Slope of BC ∴ AD ∥ BC Hence, AB II CD & AD II BC Since both pairs of opposite sides of ABCD are parallel. Hence, ABCD is a parallelogram

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.