If the axes are translated to the orthocentre | vedclass

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Question

(A) The vertices of the triangle are $A(7,5), B(-5,-7), C(7,-7)$.
Since $AC$ is vertical $(x=7)$ and $BC$ is horizontal $(y=-7)$,the triangle is a ri

Vedclass · Accepted Answer

$(6\sqrt{2}-12, 12-6\sqrt{2})$

Vedclass · Answer

(A) The vertices of the triangle are $A(7,5), B(-5,-7), C(7,-7)$.
Since $AC$ is vertical $(x=7)$ and $BC$ is horizontal $(y=-7)$,the triangle is a right-angled triangle at $C(7,-7)$.
The orthocentre $H$ of a right-angled triangle is the vertex at the right angle,so $H = (7,-7)$.
The axes are translated to $(7,-7)$,so the new coordinates $(X, Y)$ are related to the old coordinates $(x, y)$ by $X = x - 7$ and $Y = y + 7$.
The side lengths are $a = BC = 12$,$b = AC = 12$,and $c = AB = \sqrt{(7 - (-5))^2 + (5 - (-7))^2} = \sqrt{12^2 + 12^2} = 12\sqrt{2}$.
The incentre $I(x_i, y_i)$ is given by $(\frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c})$.
$x_i = \frac{12(7) + 12(-5) + 12\sqrt{2}(7)}{12 + 12 + 12\sqrt{2}} = \frac{24 + 84\sqrt{2}}{24 + 12\sqrt{2}} = \frac{2 + 7\sqrt{2}}{2 + \sqrt{2}} = \frac{(2 + 7\sqrt{2})(2 - \sqrt{2})}{4 - 2} = \frac{4 - 2\sqrt{2} + 14\sqrt{2} - 14}{2} = \frac{12\sqrt{2} - 10}{2} = 6\sqrt{2} - 5$.
$y_i = \frac{12(5) + 12(-7) + 12\sqrt{2}(-7)}{12 + 12 + 12\sqrt{2}} = \frac{60 - 84 - 84\sqrt{2}}{24 + 12\sqrt{2}} = \frac{-24 - 84\sqrt{2}}{24 + 12\sqrt{2}} = \frac{-2 - 7\sqrt{2}}{2 + \sqrt{2}} = \frac{(-2 - 7\sqrt{2})(2 - \sqrt{2})}{4 - 2} = \frac{-4 + 2\sqrt{2} - 14\sqrt{2} + 14}{2} = \frac{10 - 12\sqrt{2}}{2} = 5 - 6\sqrt{2}$.
In the new system,$X_i = x_i - 7 = 6\sqrt{2} - 5 - 7 = 6\sqrt{2} - 12$ and $Y_i = y_i - (-7) = 5 - 6\sqrt{2} + 7 = 12 - 6\sqrt{2}$.

If the axes are translated to the orthocentre of the triangle formed by the points $A(7,5), B(-5,-7), C(7,-7)$,then the coordinates of the incentre of the triangle in the new system are

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