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Ex 7.4, 4 The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices. The two given vertices of the square ABCD are A(−1, 2) and C(3, 2). Let B(x, y) be the unknown vertex. We know, all the sides of square are equal. AB = CB Using Distance Formula, AB = √((𝑥+1)^2+(𝑦−2)^2 ) CB = √((𝑥−3)^2+(𝑦−2)^2 ) Now, AB = CB √((𝑥+1)^2+(𝑦−2)^2 )=√((𝑥−3)^2+(𝑦−2)^2 ) Cancelling the square roots (𝑥+1)^2+(𝑦−2)^2= (𝑥−3)^2+(𝑦−2)^2 (𝑥+1)^2= (𝑥−3)^2+(𝒚−𝟐)^𝟐−(𝒚−𝟐)^𝟐 (𝑥+1)^2=(𝑥−3)^2 (𝑥+1)^2−(𝑥−3)^2=0 𝑥^2+1+2𝑥−(𝑥^2+9−6𝑥)=0 𝑥^2+1+2𝑥−𝑥^2−9+6𝑥=0 8𝑥−8=0 8𝑥=8 𝑥=8/8 𝑥=1 Join AC. Since all angles of square is 90° Δ ABC is a right angled triangle Applying Pythagoras theorem, 〖𝐴𝐶〗^2=𝐴𝐵^2+𝐵𝐶^2 [√((3+1)^2+(2−2)^2 )]^2 =[√((𝑥+1)^2+(𝑦−2)^2 )]^2+ [√((𝑥−3)^2+(𝑦−2)^2 )]^2 〖(4)〗^2+(0)^2=(𝑥+1)^2+(𝑦−2)^2+(𝑥−3)^2+(𝑦−2)^2 16 = 𝑥^2+1+2𝑥+𝑦^2+4−4𝑦+𝑥^2+9−6𝑥+𝑦^2+4−4𝑦 16 = 〖2𝑥〗^2−4𝑥+2𝑦^2−8𝑦+18 Putting x = 1 16 = 〖2(1)〗^2−4(1)+2𝑦^2−8𝑦+18 16 = 2 − 4 + 〖2𝑦〗^2−8𝑦+18 16 = 16 + 〖2𝑦〗^2−8𝑦 16 − 16 = 〖2𝑦〗^2−8𝑦 0 = 〖2𝑦〗^2−8𝑦 0 = 2𝑦 (𝑦−4) 𝑦 (𝑦−4) = 0 So, 𝑦=0 and 𝑦 −4=0 So, 𝑦=0 and 𝑦=4 Thus, the other two vertices are (1, 0) and (1, 4). The Square will look like this.

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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.