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If `m_1` and `m_2` are the roots of the equation `x^2+(sqrt(3)+2)x+sqrt(3)-1=0`, then the area of the `Delta` formed by lines `y=m_1x`, `y=m_2x`, `y=c` is:a.`((sqrt(33)+sqrt(11))/4)c^2`b.`((sqrt(32)+sqrt(11))/16)c`c.`((sqrt(33)+sqrt(10))/4)c^2`d.`((sqrt(33)+sqrt(21))/4)c^3` |
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Answer» `(c/m_1,c),(c/m_2,c)` `b=|c/m_1-c/m_2|=c|1/m_11/m_2|=c|(m_1-m_2)/(m_1m_2)|` `=(csqrt(m_1-m_2)^2)/|m_1m_2|` `=c{(m_1+m_2)^2-m_1m_2}/|m_1m_2|` `=(csqrt((sqrt3+2)^2-4(sqrt3-1))/(sqrt3-1))` `=c*sqrt(3^2+2^2+4sqrt3-4sqrt3+4)/(sqrt3-1)` `=c*sqrt11/(sqrt3-1)` h=c Area of triangle=`1/2*b*h=1/2*c*(sqrt33+sqrt11)/2*c` `=(sqrt33+sqrt11)/4*c^2` option a is correct. |
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