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| 1. |
if two vertices of an equilateral trangle be (0,0),(3,root 3)find 3rd vertics |
| Answer» OA = OB = AB{tex} \\Rightarrow {/tex}\xa0OA2 = OB2 = AB2\xa0\xa0OA2 =OB2{tex} \\Rightarrow {/tex}\xa0x2 + y2 = 12 .... (i)OB2 = AB2{tex} \\Rightarrow 3x + \\sqrt 3 y = 6{/tex}\xa0..... (ii){tex} \\Rightarrow y = \\frac{{6 - 3x}}{{\\sqrt 3 }}{/tex}Put the value of y in eq. (i),{tex}{x^2} + {\\left( {\\frac{{6 - 3x}}{{\\sqrt 3 }}} \\right)^2} = 12{/tex}{tex} \\Rightarrow {/tex}\xa0x = 0 or x = 3When x = 0,\xa0{tex}y = 2\\sqrt 3 {/tex}{tex}\\left( {0,2\\sqrt 3 } \\right){/tex}When x = 3y,\xa0{tex}y = - \\sqrt 3 {/tex}{tex}\\left( {3, - \\sqrt 3 } \\right){/tex} | |