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

circle $x^{2} + y^{2} = 10$ is

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Solution

The correct option is C
5

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

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

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

$P Q = 2 Q L = 2 \sqrt{O Q^{2} - O L^{2}} = 2 \sqrt{10 - 5} = 2 \sqrt{5}$

Thus area of \triangle OPQ = $\frac{1}{2} \times P Q \times O L = \frac{1}{2} \times 2 \sqrt{5} \times \sqrt{5} = 5$ .

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The area of the triangle formed by joining the origin to the points of intersection of the line x√5+2y=3√5 and circle x2+y2=10 is

The correct option is C 5 Length of perpendicular from origin to the line x\sqrt{5} + 2y = 3\sqrt{5} is OL = 3√5(√5)2+22=3√5√9=√5 Radius of the given circle = √10 = OQ = OP PQ=2QL=2√OQ2−OL2=2√10−5=2√5 Thus area of \triangle OPQ = 12×PQ×OL=12×2√5×√5=5.

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

circle $x^{2} + y^{2} = 10$ is