Lines L_1: y-x=0 and L_2: 2x+y=0 intersect th | vedclass

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(A) The lines $L_1: y=x$ and $L_2: y=-2x$ intersect at the origin $O(0,0)$.Line $L_3$ is $y=-2$.For $P$,substitute $y=-2$ in $L_1$: $-2-x=0 \implies x

Vedclass · Accepted Answer

$Statement-1$ is True,$Statement-2$ is True; $Statement-2$ is a correct explanation for $Statement-1$.

Vedclass · Answer

(A) The lines $L_1: y=x$ and $L_2: y=-2x$ intersect at the origin $O(0,0)$.Line $L_3$ is $y=-2$.For $P$,substitute $y=-2$ in $L_1$: $-2-x=0 \implies x=-2$. So $P=(-2,-2)$.For $Q$,substitute $y=-2$ in $L_2$: $2x-2=0 \implies x=1$. So $Q=(1,-2)$.The distance $OP = \sqrt{(-2-0)^2 + (-2-0)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.The distance $OQ = \sqrt{(1-0)^2 + (-2-0)^2} = \sqrt{1+4} = \sqrt{5}$.By the Angle Bisector Theorem in $	riangle OPQ$,the bisector of $\angle POQ$ divides the opposite side $PQ$ in the ratio of the sides $OP$ and $OQ$.Therefore,$\frac{PR}{RQ} = \frac{OP}{OQ} = \frac{2\sqrt{2}}{\sqrt{5}}$.$Statement-1$ is True.$Statement-2$ is the Angle Bisector Theorem,which is a standard geometric property. Thus,$Statement-2$ is True and is the correct explanation for $Statement-1$.

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