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(A) Let the two lines be $L_1$ and $L_2$ passing through the origin $O(0,0)$. Since they form an isosceles right-angled triangle with the line $2x + 3

Vedclass · Accepted Answer

$\frac{36}{13}$

Vedclass · Answer

(A) Let the two lines be $L_1$ and $L_2$ passing through the origin $O(0,0)$. Since they form an isosceles right-angled triangle with the line $2x + 3y = 6$,let the vertices be $O(0,0)$,$A$,and $B$ such that $OA = OB = r$ and $\angle AOB = 90^\circ$.Let the coordinates of $A$ be $(r \cos 	heta, r \sin 	heta)$. Since $\angle AOB = 90^\circ$,the coordinates of $B$ are $(r \cos(	heta + 90^\circ), r \sin(	heta + 90^\circ)) = (-r \sin 	heta, r \cos 	heta)$.Since $A$ and $B$ lie on the line $2x + 3y = 6$,we have:$2(r \cos 	heta) + 3(r \sin 	heta) = 6$  --- $(1)$$2(-r \sin 	heta) + 3(r \cos 	heta) = 6$  --- $(2)$Equating $(1)$ and $(2)$:$2r \cos 	heta + 3r \sin 	heta = 3r \cos 	heta - 2r \sin 	heta$$5r \sin 	heta = r \cos 	heta \Rightarrow 	an 	heta = \frac{1}{5}$Then $\sin 	heta = \frac{1}{\sqrt{26}}$ and $\cos 	heta = \frac{5}{\sqrt{26}}$.Substituting into $(1)$:$r \left(2 \cdot \frac{5}{\sqrt{26}} + 3 \cdot \frac{1}{\sqrt{26}}ight) = 6$$r \left(\frac{13}{\sqrt{26}}ight) = 6 \Rightarrow r = \frac{6 \sqrt{26}}{13}$The area of the right-angled triangle $OAB$ is $\frac{1}{2} 	imes OA 	imes OB = \frac{1}{2} r^2$.Area $= \frac{1}{2} \left(\frac{6 \sqrt{26}}{13}ight)^2 = \frac{1}{2} \cdot \frac{36 \cdot 26}{169} = \frac{18 \cdot 26}{169} = \frac{18 \cdot 2}{13} = \frac{36}{13}$ sq. units.

$A$ pair of straight lines drawn through the origin forms an isosceles triangle right-angled at the origin with the line $2x + 3y = 6$. The area (in sq. units) of the triangle so formed is

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