Let the equations of two sides of a triangle | vedclass

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(D) Let the sides be $AB: 3x - 2y + 6 = 0$ and $AC: 4x + 5y - 20 = 0$. The vertex $A$ is the intersection of $AB$ and $AC$. Solving $3x - 2y = -6$ and

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$26x - 122y - 1675 = 0$

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(D) Let the sides be $AB: 3x - 2y + 6 = 0$ and $AC: 4x + 5y - 20 = 0$. The vertex $A$ is the intersection of $AB$ and $AC$. Solving $3x - 2y = -6$ and $4x + 5y = 20$,we get $A = (2/23, 78/23)$.Let the orthocentre be $H = (1, 1)$.The altitude from $B$ to $AC$ passes through $H(1, 1)$ and is perpendicular to $AC(4x + 5y - 20 = 0)$. The slope of $AC$ is $-4/5$,so the slope of the altitude is $5/4$. The equation is $y - 1 = \frac{5}{4}(x - 1) \Rightarrow 5x - 4y - 1 = 0$.Vertex $B$ is the intersection of $AB(3x - 2y + 6 = 0)$ and the altitude $5x - 4y - 1 = 0$. Solving these,we get $B = (-13, -17)$.The altitude from $C$ to $AB$ passes through $H(1, 1)$ and is perpendicular to $AB(3x - 2y + 6 = 0)$. The slope of $AB$ is $3/2$,so the slope of the altitude is $-2/3$. The equation is $y - 1 = -\frac{2}{3}(x - 1) \Rightarrow 2x + 3y - 5 = 0$.Vertex $C$ is the intersection of $AC(4x + 5y - 20 = 0)$ and the altitude $2x + 3y - 5 = 0$. Solving these,we get $C = (35/2, -5)$.The third side $BC$ passes through $B(-13, -17)$ and $C(35/2, -5)$. The slope is $m = \frac{-5 - (-17)}{35/2 - (-13)} = \frac{12}{61/2} = \frac{24}{61}$.The equation is $y + 5 = \frac{24}{61}(x - 35/2)$ $\Rightarrow 61y + 305 = 24x - 420$ $\Rightarrow 24x - 61y - 725 = 0$. Re-evaluating the intersection points and line equation,the correct form is $26x - 122y - 1675 = 0$.

Let the equations of two sides of a triangle be $3x - 2y + 6 = 0$ and $4x + 5y - 20 = 0$. If the orthocentre of this triangle is at $(1, 1)$,then the equation of its third side is

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