Let k be an integer such that the triangle wi | vedclass

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(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_

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$\left( 2, \frac{1}{2} \right)$

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(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 28$.Substituting the vertices $(k, -3k), (5, k), (-k, 2)$:$\frac{1}{2} |k(k - 2) + 5(2 - (-3k)) + (-k)(-3k - k)| = 28$$\frac{1}{2} |k^2 - 2k + 10 + 15k + 4k^2| = 28$$|5k^2 + 13k + 10| = 56$Case $1$: $5k^2 + 13k + 10 = 56$ $\Rightarrow 5k^2 + 13k - 46 = 0$ $\Rightarrow (k - 2)(5k + 23) = 0$. Since $k$ is an integer,$k = 2$.Case $2$: $5k^2 + 13k + 10 = -56 \Rightarrow 5k^2 + 13k + 66 = 0$. The discriminant $D = 13^2 - 4(5)(66) = 169 - 1320 For $k = 2$,the vertices are $A(2, -6), B(5, 2), C(-2, 2)$.The side $BC$ is horizontal $(y = 2)$,so the altitude from $A$ is the vertical line $x = 2$.The slope of $AC$ is $m_{AC} = \frac{2 - (-6)}{-2 - 2} = \frac{8}{-4} = -2$. The altitude from $B$ to $AC$ has slope $\frac{1}{2}$ and passes through $B(5, 2)$.Equation of altitude from $B$: $y - 2 = \frac{1}{2}(x - 5)$ $\Rightarrow 2y - 4 = x - 5$ $\Rightarrow x - 2y = 1$.Since the orthocentre $H$ lies on $x = 2$,substituting $x = 2$ into $x - 2y = 1$ gives $2 - 2y = 1$ $\Rightarrow 2y = 1$ $\Rightarrow y = \frac{1}{2}$.Thus,the orthocentre is $\left( 2, \frac{1}{2} \right)$.

Let $k$ be an integer such that the triangle with vertices $(k, -3k)$,$(5, k)$,and $(-k, 2)$ has an area of $28$ sq. units. Then the orthocentre of this triangle is at the point:

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