The lines L_1: y-x=0 and L_2: 2x+y=0 intersec | vedclass

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(A) Given lines are $L_1: y-x=0$,$L_2: 2x+y=0$,and $L_3: y+2=0$.The intersection of $L_1$ and $L_3$ is found by substituting $y=-2$ into $y-x=0$,givin

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Statement-$1$ is true,Statement-$2$ is false

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(A) Given lines are $L_1: y-x=0$,$L_2: 2x+y=0$,and $L_3: y+2=0$.The intersection of $L_1$ and $L_3$ is found by substituting $y=-2$ into $y-x=0$,giving $x=-2$. So,$P = (-2, -2)$.The intersection of $L_2$ and $L_3$ is found by substituting $y=-2$ into $2x+y=0$,giving $2x-2=0$,so $x=1$. So,$Q = (1, -2)$.The origin $O$ is $(0, 0)$. The lengths are $OP = \sqrt{(-2-0)^2 + (-2-0)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$ and $OQ = \sqrt{(1-0)^2 + (-2-0)^2} = \sqrt{1+4} = \sqrt{5}$.By the Angle Bisector Theorem in $	riangle OPQ$,the bisector of $\angle POQ$ divides the opposite side $PQ$ in the ratio of the adjacent sides $OP$ and $OQ$.Thus,$PR : RQ = OP : OQ = 2\sqrt{2} : \sqrt{5}$. Hence,Statement-$1$ is true.Statement-$2$ is false because the angle bisector of a triangle does not necessarily divide it into two similar triangles (unless the triangle is isosceles with respect to that angle).Therefore,Statement-$1$ is true and Statement-$2$ is false.

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