Question

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

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is C $5$

Length of perpendicular from origin to the line $x\sqrt{5}+2y=3\sqrt{5}$ is

$OL=\frac{3\sqrt{5}}{\sqrt{{\left(\sqrt{5}\right)}^2+2^2}}=\frac{3\sqrt{5}}{\sqrt{9}}=\sqrt{5}$

Radius of the given circle $=\sqrt{10}=OQ=OP$

$PQ=2QL=2\sqrt{{QL}^2-{OL}^2}=2\sqrt{10-5}=2\sqrt{5}$

Thus, area of$△OPQ=\frac{1}{2}\times PQ\times OL$

area of$△OPQ$$=\frac{1}{2}\times 2\sqrt{5}\times \sqrt{5}=5sq.units$