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

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(C) 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-angle

Vedclass · Accepted Answer

$D, A, F, C$

Vedclass · Answer

(C) 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 $B(-6,3)$.$1$. $P$ is the midpoint of $BC$. $P = (\frac{-6-6}{2}, \frac{3-3}{2}) = (-6,0)$. Thus,$i ightarrow D$.$2$. 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 $y - 3 = \frac{1}{2}(x - 6)$ $\Rightarrow 2y - 6 = x - 6$ $\Rightarrow x = 2y$. The $x$-axis intersection $(y=0)$ is $Q(0,0)$. Thus,$ii ightarrow A$.$3$. The orthocentre $R$ of a right-angled triangle is the vertex where the right angle is located. Here,$R = B(-6,3)$. Thus,$iii ightarrow F$.$4$. The centroid $S$ is $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}) = (\frac{6-6-6}{3}, \frac{3+3-3}{3}) = (-2,1)$. Thus,$iv ightarrow C$.The correct sequence is $D, A, F, C$.

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

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