Question

Show that area of the triangle formed by the lines$\mathrm{y}={\mathrm{m}}_1\mathrm{x}$, $\mathrm{y}={\mathrm{m}}_2\mathrm{x}$, and $\mathrm{y}=\mathrm{c}$ is equal to $\frac{{\mathrm{c}}^2(\surd 33+\surd 11)}{4}$, where ${\mathrm{m}}_1$and ${\mathrm{m}}_2$ are the roots of the equation ${\mathrm{x}}^2+(\surd 3+2)\mathrm{x}+\surd 3–1=0$.

Answer 1

Given that, ${\mathrm{m}}_1$and ${\mathrm{m}}_2$ are the roots of the quadratic equation ${\mathrm{x}}^2+(\surd 3+2)\mathrm{x}+\surd 3–1=0$.

Comparing ${\mathrm{x}}^2+(\surd 3+2)\mathrm{x}+\surd 3–1=0$ with the standard form of quadratic equation ${\mathrm{ax}}^2$$-$(sum of the roots${={\mathrm{m}}_1+{\mathrm{m}}_2}_{}$)$\mathrm{x}$ $+$product of the roots(${\mathrm{m}}_1{\mathrm{m}}_2$$=0$, we get

${\mathrm{m}}_1+{\mathrm{m}}_2=$$-\surd 3+2$ …………………………. .$\left(1\right)$

${\mathrm{m}}_1{\mathrm{m}}_2$ $=$ $\surd 3–1$

${({\mathrm{m}}_1-{\mathrm{m}}_2)}^2=$${({\mathrm{m}}_1+{\mathrm{m}}_2)}^2-4{\mathrm{m}}_1{\mathrm{m}}_2$

$=$ ${(\surd 3+2)}^2-4(\surd 3–1)$

$=$${(\surd 3)}^2+{\left(2\right)}^2+2\times 2\times \surd 3-4\surd 3+4$

$=$$3+4+4\surd 3-4\surd 3+4$

${({\mathrm{m}}_1-{\mathrm{m}}_2)}^2=$ $11$

${\mathrm{m}}_1-{\mathrm{m}}_2=\surd 11$……………………………$\left(2\right)$

Adding equation $\left(1\right)$ and equation $\left(2\right)$, we get

$2{\mathrm{m}}_1$ $=\surd 3+2+\surd 11$

⇒ ${\mathrm{m}}_1$ $=\frac{\surd 3+2+\surd 11}{2}$

Similarly, ${\mathrm{m}}_2$ $=\surd 3+2-\frac{\surd 3+2+\surd 11}{2}$

$=\frac{2\surd 3+4-\surd 3-2-\surd 11}{2}$

⇒ ${\mathrm{m}}_2$$=\frac{\surd 3+2-\surd 11}{2}$

Given equations of line are:

$\mathrm{y}={\mathrm{m}}_1\mathrm{x}$…………….$\left(3\right)$

$\mathrm{y}={\mathrm{m}}_2\mathrm{x}$…………….$\left(4\right)$

$\mathrm{y}=\mathrm{c}$…………………$\left(5\right)$

We will solve the above equations to obtain the coordinates of the triangle

On solving eq $\left(3\right)$and $\left(4\right)$, we get

we get coordinate of intersecting point $\mathrm{A}$$(0,0)$

On solving eq $\left(4\right)$ and$\left(5\right)$, we get

coordinate of intersecting point $\mathrm{B}$$(\frac{\mathrm{c}}{{\mathrm{m}}_2},\mathrm{c})$

On solving eq $\left(3\right)$ and $\left(5\right)$, we get

we get coordinate $\mathrm{C}(\frac{\mathrm{c}}{\mathrm{m}1},\mathrm{c})$

Now the area of triangle when three of its vertices are given $=\frac{1}{2}\left[\right(x_1({\mathrm{y}}_2–{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3–{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1–{\mathrm{y}}_2)]$

⇒ Area $=\frac{1}{2}\left[\right({\mathrm{x}}_1({\mathrm{y}}_2–{\mathrm{y}}_3)+{\mathrm{x}}_2({\mathrm{y}}_3–{\mathrm{y}}_1)+{\mathrm{x}}_3({\mathrm{y}}_1–{\mathrm{y}}_2)]$

$=\frac{1}{2}\left[\right(0(\mathrm{c}–\mathrm{c})+\frac{\mathrm{c}}{{\mathrm{m}}_2}(\mathrm{c}–0)+\frac{\mathrm{c}}{{\mathrm{m}}_1}(0–\mathrm{c})]$

$=\frac{1}{2}[\frac{{\mathrm{c}}^2}{{\mathrm{m}}_2}-\frac{{\mathrm{c}}^2}{{\mathrm{m}}_1}]$

$=\frac{{\mathrm{c}}^2}{2}[\frac{1}{{\mathrm{m}}_2}-\frac{1}{{\mathrm{m}}_1}]$

⇒ Area $=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{{\mathrm{m}}_1{\mathrm{m}}_2}\right]$……………..$\left(6\right)$

Now substitute the value of ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$ in equation $\left(6\right)$, we get

Area$=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\displaystyle \frac{\surd 3+2+\surd 11}{2}}-\frac{\surd 3+2-\surd 11}{2}}{\left(\frac{\surd 3+2+\surd 11}{2}\right)\left(\frac{\surd 3+2-\surd 11}{2}\right)}\right]$

$=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\displaystyle \frac{\surd 3+2+\surd 11-\surd 3-2+\surd 11}{2}}}{\frac{{(\surd 3+2)}^2-{(\surd 11)}^2}{4}}\right]$

$=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\displaystyle \frac{2\surd 11}{2}}}{\frac{{{(\surd 3)}^2+\left(2\right)}^2+2\times \surd 3\times 2-11}{4}}\right]$

$=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\displaystyle \surd 11}}{\frac{3+4+4\surd 3-11}{4}}\right]$

$=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\displaystyle \surd 11}}{\frac{4\surd 3-4}{4}}\right]$

$=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\displaystyle \surd 11}}{\frac{4(\surd 3-1)}{4}}\right]$

$=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\displaystyle \surd 11}}{(\surd 3-1}\right]$

$=\frac{{\mathrm{c}}^2}{2}[\frac{{\displaystyle \surd 11}}{(\surd 3-1)}\times \frac{(\surd 3+1)}{(\surd 3+1)}]$

$=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\displaystyle \surd 33+\surd 11}}{{(\surd 3)}^2-{\left(1\right)}^2}\right]$

$=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\displaystyle \surd 33+\surd 11}}{3-1}\right]$

$=\frac{{\mathrm{c}}^2}{2}\left[\frac{{\displaystyle \surd 33+\surd 11}}{2}\right]$

$=\frac{{\mathrm{c}}^2(\surd 33+\surd 11)}{4}$

Thus, area of the triangle formed by the lines$\mathrm{y}={\mathrm{m}}_1\mathrm{x}$, $\mathrm{y}={\mathrm{m}}_2\mathrm{x}$, and $\mathrm{y}=\mathrm{c}$ is equal to $\frac{{\mathrm{c}}^2(\surd 33+\surd 11)}{4}$. Hence proved