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

Question

TestHub · Accepted Answer

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

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

Radius of the given circle $= \sqrt{10} = O Q = O P$

$P Q = 2 Q L$

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

$= 2 \sqrt{10 - 5} = 2 \sqrt{5}$

Thus, area of $Δ O P Q = \frac{1}{2} \times P Q \times O L$

$= \frac{1}{2} \times 2 \sqrt{5} \times \sqrt{5} = 5$ sq units

Hence, option (c) is correct.

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