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The three vertices of a parallelogram are (3,4), (3,8), and (9,8). Find the fourth vertex

Answer» Let A(3, 4), B(3, 8) and C(9, 8) be the given points.Let the fourth vertex be D (x, y){tex}A B = \\sqrt { ( 3 - 3 ) ^ { 2 } + ( 8 - 4 ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad A B = \\sqrt { 0 + ( 4 ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad A B = \\sqrt { 16 }{/tex}{tex}\\Rightarrow{/tex}\xa0AB = 4{tex}B C = \\sqrt { ( 9 - 3 ) ^ { 2 } + ( 8 - 8 ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad B C = \\sqrt { ( 6 ) ^ { 2 } + 0 }{/tex}{tex}\\Rightarrow \\quad B C = \\sqrt { 36 }{/tex}{tex}\\Rightarrow{/tex}\xa0BC = 6{tex}C D = \\sqrt { ( x - 9 ) ^ { 2 } + ( y - 8 ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad C D = \\sqrt { x ^ { 2 } + \\left( 9 ^ { 2 } \\right) - 18 x + y ^ { 2 } + \\left( 8 ^ { 2 } \\right) - 16 y }{/tex}{tex}\\Rightarrow \\quad C D = \\sqrt { x ^ { 2 } + 81 - 18 x + y ^ { 2 } + 64 - 16 y }{/tex}{tex}\\Rightarrow \\quad C D = \\sqrt { x ^ { 2 } - 18 x + y ^ { 2 } - 16 y + 145 }{/tex}{tex}A D = \\sqrt { ( x - 3 ) ^ { 2 } + ( y - 4 ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad A D = \\sqrt { x ^ { 2 } + 9 - 6 x + y ^ { 2 } + 16 - 8 y }{/tex}{tex}\\Rightarrow \\quad A D = \\sqrt { x ^ { 2 } - 6 x + y ^ { 2 } - 8 y + 25 }{/tex}Since ABCD is a parallelogram and opposite sides of a parallelogram are equal.AB = CD and AD = BCAB = CDAB2 = CD2{tex}\\Rightarrow{/tex}\xa0x2 - 18x + y2 -16y + 145 = 16{tex}\\Rightarrow{/tex}\xa0x2 - 18x + y2 - 16y + 145 - 16 = 0{tex}\\Rightarrow{/tex}\xa0x2 - 18x + y2 - 16y + 129 = 0 ...(i)BC = ADBC2 = AD2x2 - 6x + y2 - 8y + 25 = 36{tex}\\Rightarrow{/tex}\xa0x2 - 6x + y2 - 8y + 25 -36 = 0{tex}\\Rightarrow{/tex}\xa0x2 - 6x + y2 - 8y - 11 = 0 ...(ii)Solving (i) and (ii), we getx = 9, y = 4{tex}\\therefore{/tex}\xa0The fourth vertex is D(9, 4).


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