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(C) Let the perpendicular lines through the origin be $L_1$ and $L_2$. Since they are perpendicular and form an isosceles triangle with the line $2x +

Vedclass · Accepted Answer

$\frac{36}{13}$

Vedclass · Answer

(C) Let the perpendicular lines through the origin be $L_1$ and $L_2$. Since they are perpendicular and form an isosceles triangle with the line $2x + 3y = 6$,the triangle $	riangle OAB$ is a right-angled isosceles triangle with $\angle AOB = 90^\circ$ and $OA = OB$.Let $OP$ be the perpendicular distance from the origin $(0, 0)$ to the line $2x + 3y - 6 = 0$.$OP = \frac{|2(0) + 3(0) - 6|}{\sqrt{2^2 + 3^2}} = \frac{6}{\sqrt{4 + 9}} = \frac{6}{\sqrt{13}}$.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,$OP = AP = BP = \frac{6}{\sqrt{13}}$.The base of the triangle $AB = AP + BP = 2 	imes OP = 2 	imes \frac{6}{\sqrt{13}} = \frac{12}{\sqrt{13}}$.The area of $	riangle OAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes OP$.Area $= \frac{1}{2} 	imes \left(\frac{12}{\sqrt{13}}ight) 	imes \left(\frac{6}{\sqrt{13}}ight) = \frac{36}{13}$ sq units.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

If a pair of perpendicular lines through the origin together with the straight line $2x + 3y = 6$ form an isosceles triangle,then the area of that triangle (in sq units) is

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