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The points A(1,-2) B(2,3) C(k,-2) D(-4,-3) are the vertices of a parallelogram. Find value of k |
| Answer» A(1, -2), B(2, 3), C(a, 2) and D(-4, -3)are the vertices of a parallelogram.{tex}\\Rightarrow{/tex}AB = CD and AD = BCConsider AD = BC{tex}\\Rightarrow{/tex}AD2 = BC2{tex}\\Rightarrow{/tex}(-4 -1)2 + (-3 + 2)2 = (a - 2)2 + (2 - 3)2{tex}\\Rightarrow{/tex}(-5)2 = (a - 2)2{tex}\\Rightarrow{/tex}\xa0a - 2 = -5{tex}\\Rightarrow{/tex}\xa0a = -3Area of\xa0{tex}\\Delta A B C = \\frac { 1 } { 2 } | 1 ( 3 - 2 ) + 2 ( 2 + 2 ) - 3 ( - 2 - 3 ) |{/tex}{tex}= \\frac { 1 } { 2 } | 1 + 8 + 15 |{/tex}{tex}= \\frac { 1 } { 2 } \\times 24{/tex}= 12 sq. units.Diagonal of a parallelogram divides it into two equal triangles.Area of parallelogram ABCD = 2\xa0{tex}\\times{/tex}12=24 sq. units{tex}\\Rightarrow {/tex}\xa0AB\xa0{tex}\\times{/tex}Height = 24{tex}\\Rightarrow {/tex}\xa0AB{tex}\\times{/tex}Height = 24{tex}\\Rightarrow \\left[ \\sqrt { ( 2 - 1 ) ^ { 2 } + ( 3 + 2 ) ^ { 2 } } \\right] \\times \\text { Height } = 24{/tex}{tex}\\Rightarrow [ \\sqrt { 1 + 25 } ] \\times \\text { Height } = 24{/tex}{tex}\\Rightarrow \\text { Height } = \\frac { 24 } { \\sqrt { 26 } } = \\frac { 12 \\times \\sqrt { 2 } \\times \\sqrt { 2 } } { \\sqrt { 13 } \\times \\sqrt { 2 } }{/tex}{tex}= \\frac { 12 \\sqrt { 2 } } { \\sqrt { 13 } }{/tex} | |