In the triangle with vertices at A(6, 3), B(- | vedclass

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Question

(A) The vertices are $A(6, 3), B(-6, 3)$,and $C(-6, -3)$.Since $AB$ is horizontal $(y=3)$ and $BC$ is vertical $(x=-6)$,$	riangle ABC$ is a right-ang

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$D, A, F, C$

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(A) The vertices are $A(6, 3), B(-6, 3)$,and $C(-6, -3)$.Since $AB$ is horizontal $(y=3)$ and $BC$ is vertical $(x=-6)$,$	riangle ABC$ is a right-angled triangle with the right angle at $B(-6, 3)$.$P$ is the midpoint of $BC$: $P = (\frac{-6-6}{2}, \frac{3-3}{2}) = (-6, 0)$. Thus $(i) ightarrow D$.For $Q$,the line $AC$ passes through $(6, 3)$ and $(-6, -3)$. The slope $m = \frac{-3-3}{-6-6} = \frac{-6}{-12} = \frac{1}{2}$. The equation is $y - 3 = \frac{1}{2}(x - 6) \Rightarrow 2y - 6 = x - 6 \Rightarrow x = 2y$. At $y=0$,$x=0$. So $Q = (0, 0)$. Thus $(ii) ightarrow A$.$R$ is the orthocentre. In a right-angled triangle,the orthocentre is the vertex at the right angle,which is $B(-6, 3)$. Thus $(iii) ightarrow F$.$S$ is the centroid: $S = (\frac{6-6-6}{3}, \frac{3+3-3}{3}) = (-2, 1)$. Thus $(iv) ightarrow C$.The correct sequence is $D, A, F, C$.

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

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