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(A) Let the triangle be $	riangle ABC$ with the right angle at the origin $A(0, 0)$. The base of the triangle lies on the line $2x + 3y = 6$.Since th

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$\frac{36}{13}$

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(A) Let the triangle be $	riangle ABC$ with the right angle at the origin $A(0, 0)$. The base of the triangle lies on the line $2x + 3y = 6$.Since the triangle is a right-angled isosceles triangle with the right angle at the origin,the two lines passing through the origin make an angle of $45^{\circ}$ with the line $2x + 3y = 6$.The perpendicular distance $p$ from the origin $A(0, 0)$ to the line $2x + 3y - 6 = 0$ is given by:$p = \frac{|2(0) + 3(0) - 6|}{\sqrt{2^2 + 3^2}} = \frac{6}{\sqrt{13}}$.In a right-angled isosceles triangle,the altitude from the right-angle vertex to the hypotenuse bisects the hypotenuse and is equal to half the length of the hypotenuse.Thus,the length of the altitude $AL = p = \frac{6}{\sqrt{13}}$.The base $BC$ of the triangle is $2p$ because the altitude $AL$ divides the triangle into two smaller right-angled isosceles triangles with legs of length $p$.Area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2p) 	imes p = p^2$.Area $= \left(\frac{6}{\sqrt{13}}ight)^2 = \frac{36}{13}$ sq. units.

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

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