Question

The area of the triangle formed by joining the origin to the points of intersection of the lines $x\sqrt{5}+2y=3\sqrt{5}$ and circle $x^2+y^2=10$

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is C $5$

The points of intersection of line $\sqrt{5}x+2y=3\sqrt{5}$ and the circe $x^2+y^2=10$ we have

$y=\frac{3\sqrt{5}}{2}-\frac{\sqrt{5}x}{2}$ .......$\left(1\right)$

Substituting y from eqn.$\left(1\right)$ in equation of circle , we get

$x^2+{(\frac{3\sqrt{5}}{2}-\frac{\sqrt{5}x}{2})}^2=10$ .......$\left(2\right)$

$\Rightarrow x^2+\frac{45}{4}+\frac{5x^2}{4}-2\times \frac{3\sqrt{5}}{2}\times \frac{\sqrt{5}x}{2}=10$

$\Rightarrow \frac{45}{4}+\frac{4x^2+5x^2}{4}-\frac{30x}{4}=10$

$\Rightarrow 9x^2-30x+45-40=0$

$\Rightarrow 9x^2-30x+5=0$ .......$\left(3\right)$

Roots are $x=\frac{30\pm \sqrt{900-4\times 9\times 5}}{2\times 9}=\frac{30\pm \sqrt{120}}{18}=\frac{5\pm 2\sqrt{5}}{3}$ .........$\left(4\right)$

$\therefore x=\{\frac{5-2\sqrt{5}}{3},\frac{5+2\sqrt{5}}{3}\}$

Substitute $x=\frac{5-2\sqrt{5}}{3}$ in $\left(1\right)$ we get

$y=\frac{3\sqrt{5}}{2}-\frac{\sqrt{5}\times \frac{5-2\sqrt{5}}{3}}{2}$

$y=\frac{3\sqrt{5}}{2}-\frac{5\sqrt{5}-10}{6}$

$y=\frac{9\sqrt{5}-5\sqrt{5}+10}{6}$

$y=\frac{4\sqrt{5}+10}{6}$

$y=\frac{2\sqrt{5}+5}{3}$

Substitute $x=\frac{5+2\sqrt{5}}{3}$ in $\left(1\right)$ we get

$y=\frac{3\sqrt{5}}{2}-\frac{\sqrt{5}\times \frac{5+2\sqrt{5}}{3}}{2}$

$y=\frac{3\sqrt{5}}{2}-\frac{5\sqrt{5}+10}{6}$

$y=\frac{9\sqrt{5}-5\sqrt{5}-10}{6}$

$y=\frac{4\sqrt{5}-10}{6}$

$y=\frac{2\sqrt{5}-5}{3}$

hence vertices of triangle are $A(0,0),B(\frac{5+2\sqrt{5}}{3},\frac{2\sqrt{5}-5}{3})$ and $C(\frac{5-2\sqrt{5}}{3},\frac{2\sqrt{5}+5}{3})$

Area of a triangle$=\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ \frac{5+2\sqrt{5}}{3}& \frac{2\sqrt{5}-5}{3}& 1\\ \frac{5-2\sqrt{5}}{3}& \frac{2\sqrt{5}+5}{3}& 1\end{array}\right|$

$=\frac{1}{2}[{\left(\frac{5+2\sqrt{5}}{3}\right)}^2+{\left(\frac{5-2\sqrt{5}}{3}\right)}^2]$

$=\frac{1}{2\times 9}[25+20+20\sqrt{5}+25-20\sqrt{5}+20]$

$=\frac{1}{2\times 9}\left[2(25+20)\right]$

$=\frac{45}{9}=5$sq.units.