Question

prove that area of triangle formed by lines y=m1x,y=m2x and y=c is 1/4 c^2√11 (√3+1),where m1 ,m2 are roots of equation x^2+ (√3+2)x+√3-1=0​

Answer 1

Given lines

$L_1:y=m_1x$

$L_2:y=m_2x$

$L_3:y=c$

The intersection of line $L_3$ with $L_1$ and $L_2$ will be respectively

$(\frac{c}{m_1},c)$ and $(\frac{c}{m_2},c)$

The area of the triangle formed will be

$A=\frac{1}{2}\times c\times \sqrt{(\frac{c}{m_1}-\frac{c}{m_2})^2+(c-c)^2}$

$\implies A=\frac{1}{2}\times c\times c(\frac{1}{m_1}-\frac{1}{m_2})$

$\implies A=\frac{c^2}{2}(\frac{m_2-m_1}{m_1m_2})$

But given that $m_1$ and $m_2$ are roots of quadratic equation $x^2+(\sqrt{3}+1)x+\sqrt{3}-1=0$

Therefore,

$m_1+m_2=-(\sqrt{3}+2)$

And

$m_1m_2=\sqrt{3}-1$

Also

$(m_2-m_1)^2=(m_1+m_2)^2-4m_1m_2$

$\implies (m_2-m_1)^2=(\sqrt{3}+2)^2-4(\sqrt{3}-1)$

$\implies (m_2-m_1)^2=3+4\sqrt{3}+4-4\sqrt{3}+4$

$\implies (m_2-m_1)^2=11$

$\implies m_2-m_1=\sqrt{11}$

Therefore, the area will be

$A=\frac{c^2}{2}{\frac{\sqrt{11}}{\sqrt{3}-1}$

$\implies A=\frac{c^2}{2}{\frac{\sqrt{11}}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$

$\implies A=\frac{c^2}{2}{\frac{\sqrt{11}(\sqrt{3}+1)}{(\sqrt{3})^2-1^2}$

$\implies A=\frac{c^2}{2}{\frac{\sqrt{11}(\sqrt{3}+1)}{2}$

$\implies A=\frac{1}{4}c^2\sqrt{11}(\sqrt{3}+1)$       (Proved)

Hope this helps.

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