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Find the point on x-axis which is equidistant from (2,5) & (-2,9) |
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Answer» on x axis y= 0let the coordinates are O (x,0),A(2,5),B=(-2,9){tex}\\begin{array}{l}\\mathrm{OA}=\\mathrm{OB}\\\\\\mathrm{OA}^2=\\mathrm{OB}^2\\\\(\\mathrm x-2)^2+(0-5)^2=(\\mathrm x+2)^2+(0-9)^2\\\\\\mathrm x^2-4\\mathrm x+4+25=\\mathrm x^2+4\\mathrm x+4+81\\\\-8\\mathrm x=81-25=56\\\\\\mathrm x=-7\\\\\\mathrm{hence}\\;\\mathrm{the}\\;\\mathrm{point}\\;\\mathrm O\\;(-7,0)\\end{array}{/tex} Check exercise 7.1 of ncert same aisa question hai ncert me... apko proper solution isi app me mil jayega |
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