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(A) Let the lines be $L_1: y = x + 1$,$L_2: y = 4x - 8$,and $L_3: y = mx + c$. The orthocentre $H$ is $(3, -1)$.First,find the vertex $P$ by solving $

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(A) Let the lines be $L_1: y = x + 1$,$L_2: y = 4x - 8$,and $L_3: y = mx + c$. The orthocentre $H$ is $(3, -1)$.First,find the vertex $P$ by solving $L_1$ and $L_2$: $x + 1 = 4x - 8$ $\Rightarrow 3x = 9$ $\Rightarrow x = 3$. Then $y = 3 + 1 = 4$. So,$P = (3, 4)$.The altitude from $P$ to $QR$ passes through $H(3, -1)$. Since $P$ and $H$ both have $x = 3$,the altitude is the vertical line $x = 3$. Thus,$QR$ must be a horizontal line,so $m = 0$.The line $QR$ is $y = c$. Since $H(3, -1)$ is the orthocentre,the line $QH$ is perpendicular to $PR$. The slope of $PR$ (line $L_2$) is $4$,so the slope of $QH$ is $-1/4$.The line $QH$ passes through $H(3, -1)$ and $Q$. $Q$ is the intersection of $L_1$ and $L_3$. $y = x + 1$ and $y = c \Rightarrow x = c - 1$. So $Q = (c - 1, c)$.Slope of $QH = \frac{c - (-1)}{(c - 1) - 3} = \frac{c + 1}{c - 4} = -\frac{1}{4}$.$4c + 4 = -c + 4$ $\Rightarrow 5c = 0$ $\Rightarrow c = 0$.Thus,$m - c = 0 - 0 = 0$.

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

If the orthocentre of the triangle formed by the lines $y = x + 1$,$y = 4x - 8$,and $y = mx + c$ is at $(3, -1)$,then $m - c$ is :

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