The equations of the sides AB,BC,and CA of a | vedclass

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(C) The equations of the sides are: $AB: 2x + y = 0$ $(1)$ $BC: x + py = 15a$ $(2)$ $CA: x - y = 3$ $(3)$ Vertex $A$ is the intersection of $AB$ and $

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$3$

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(C) The equations of the sides are: $AB: 2x + y = 0$ $(1)$ $BC: x + py = 15a$ $(2)$ $CA: x - y = 3$ $(3)$ Vertex $A$ is the intersection of $AB$ and $CA$: $2x + y = 0$ and $x - y = 3$. Adding these gives $3x = 3$,so $x = 1$ and $y = -2$. Thus,$A = (1, -2)$. Since the orthocentre $H = (2, a)$ lies on the altitude from $A$ to $BC$,the slope of $AH$ is perpendicular to the slope of $BC$. Slope of $AH = \frac{a - (-2)}{2 - 1} = a + 2$. Slope of $BC = -\frac{1}{p}$. Since $AH \perp BC$,$(a + 2) 	imes (-\frac{1}{p}) = -1 \implies a + 2 = p$. Vertex $B$ is the intersection of $AB$ and $BC$: $y = -2x$ and $x + p(-2x) = 15a \implies x(1 - 2p) = 15a \implies x = \frac{15a}{1 - 2p}, y = \frac{-30a}{1 - 2p}$. The altitude from $B$ to $AC$ has slope $m = -1$ (since slope of $AC = 1$). The equation of altitude from $B$ is $y - y_B = -1(x - x_B) \implies x + y = x_B + y_B$. Substituting $H(2, a)$ into this: $2 + a = \frac{15a - 30a}{1 - 2p} = \frac{-15a}{1 - 2p}$. Using $p = a + 2$,we get $2 + a = \frac{-15a}{1 - 2(a + 2)} = \frac{-15a}{-3 - 2a} = \frac{15a}{2a + 3}$. $(a + 2)(2a + 3) = 15a \implies 2a^2 + 7a + 6 = 15a \implies 2a^2 - 8a + 6 = 0 \implies a^2 - 4a + 3 = 0$. $(a - 1)(a - 3) = 0$. Since $-\frac{1}{2} Then $p = a + 2 = 1 + 2 = 3$.

The equations of the sides $AB$,$BC$,and $CA$ of a triangle $ABC$ are $2x + y = 0$,$x + py = 15a$,and $x - y = 3$ respectively. If its orthocentre is $(2, a)$,where $-\frac{1}{2} < a < 2$,then $p$ is equal to...

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