Ex 7.1, 6

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Ex 7.1 , 6 Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) Let the points be A(−1, −2) , B(1, 0) C(−1, 2) , D(−3, 0) We find the distances AB, BC, CD & AD By using distance formula Finding AB x1 = −1 , y1 = −2 x2 = 1 , y2 = 0 AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 1 −(−1))2+(0 −(−2))2) = √((1+1)2+(0+2)2) = √((2)2+(2)2) = √(2(2)2) = 2√2 By using distance formula Finding BC x1 = 1 , y1 = 0 x2 = −1 , y2 = 2 BC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −1 −1)2+(2 −0)2) = √((−2)2+(2)2) = √((2)2+(2)2) = √(2(2)2) = 2√2 By using distance formula Finding CD x1 = −1 , y1 = 2 x2 = −3 , y2 = 0 CD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −3 −(−1))2+(0 −2)2) = √((−3+1)2+(−2)2) = √((−2)2+(−2)2) = √((2)2+(2)2) = √(2(2)2) = 2√2 By using distance formula Finding AD x1 = −1 , y1 = −2 x2 = −3 , y2 = 0 AD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −3 −(−1))2+(0 −(−2))2) = √((−3+1)2+(2)2) = √((−2)2+(2)2) = √((2)2+(2)2) = √(2(2)2) = 2√2 Since, AB = BC = CD = AD = 2√2 All sides of quadrilateral are equal, it can be a rhombus or a square. We find the length of diagonals of the quadrilaterals i.e. , AC & BD Finding AC x1 = −1 , y1 = −2 x2 = −1 , y2 = 2 AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( −1 −(−1))2+(2 −(−2))2) = √((−1+1)2+(2+ 2)2) = √((0)2+(4)2) = √((4)2) = 4 Now, Finding BD x1 = 1 , y1 = 0 x2 = −3 , y2 = 0 BD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 1 −(−3))2+(0 −0)2) = √(( 1+3)2+(0)2) = √((4)2+(0)2) = √((4)2) = 4 Since, AB = BC = CD = AD = 2√2 All sides of quadrilateral are equal,  & AC = BD = 4 Hence, both diagonals are equal Since, all sides and both diagonals are equal ABCD is a square Ex 7.1 , 6 Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4) Let the points be A(−3, 5) , B(3, 1) C(0, 3) , D(−1, −4) We find the distances AB, BC, CD & AD By using distance formula Finding AB x1 = −3 , y1 = 5 x2 = 3 , y2 = 1 AB = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 3 −(−3))2+(1 −5)2) = √((3+3)2+(−4)2) = √((6)2+(4)2) = √(36+16) = √52 Finding BC x1 = 3 , y1 = 1 x2 = 0 , y2 = 3 BC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √((0 −3)2+(3 −1)2) = √((−3)2+(2)2) = √((3)2+(2)2) = √(9+4) = √13 Finding CD x1 = 0 , y1 = 3 x2 = −1 , y2 = −4 CD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √((−1−0)2+(−4−3)2) = √((−1)2+(−7)2) = √((1)2+(7)2) = √(1+49) = √50 By using distance formula Finding AD x1 = −3 , y1 = 5 x2 = −1 , y2 = −4 AD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √((−1−(−3))2+(−4−5)2) = √((−1+3)2+(−9)2) = √((2)2+(9)2) = √(4+81) = √85 Hence, AB = √52 , BC = √13 , CD = √50 , AD = √85 Since, AB ≠ BC ≠ CD ≠ AD So, it is not a quadrilateral Why not a quadrilateral? Points are A(−3, 5) , B(3, 1) C(0, 3) , D(−1, −4) Here points A,B,C are collinear. We check if AC + BC = AB Finding AC By using distance formula Finding AC x1 = −3 , y1 = 5 x2 = 0 , y2 = 3 AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √((0−(−3))2+(3−5)2) = √((3)2+(−2)2) = √(9+4) = √13 Now, AB = √52 = √(4 ×13) = √4 × √13 = 2√13 AC = √13 BC = √13 Since AC + BC = AB Hence, points A,B,C are collinear. So, ABCD is a triangle, not a quadrilateral Ex 7.1 , 6 Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (iii) (4, 5), (7, 6), (4, 3), (1, 2) Let the points be A(4, 5) , B(7, 6) C(4, 3) , D(1, 2) We find the distances AB, BC, CD & AD By using distance formula Finding AB x1 = 4 , y1 = 5 x2 = 7 , y2 = 6 AB = √((𝑥2 −𝑥1)2+(𝑦2 −1)2) = √(( 7 −4)2+(6 −5)2) = √((3)2+(1)2) = √(9+1) = √10 Finding BC x1 = 7 , y1 = 6 x2 = 4 , y2 = 3 BC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 4 −7)2+(3 −6)2) = √((−3)2+(−3)2) = √((3)2+(3)2) = √(9+9) = √18 Finding CD x1 = 4 , y1 = 3 x2 = 1 , y2 = 2 CD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 1 −4)2+(2 −3)2) = √((−3)2+(−1)2) = √((3)2+(1)2) = √(9+1) = √10 Finding AD x1 = 4 , y1 = 5 x2 = 1 , y2 = 2 AD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 1 −4)2+(2 −5)2) = √((−3)2+(−3)2) = √((3)2+(3)2) = √(9+9) = √18 Since, AB = CD = √10 & AD = BC = √18 Since both pairs of opposite sides of quadrilateral are equal, it would be a parallelogram