Suppose triangle ABC is an isosceles triangle | vedclass

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(D) Let $C = (x, y)$. Since $	riangle ABC$ is an isosceles right-angled triangle at $C$,we have $AC = BC$ and $AC \perp BC$.From $AC^2 = BC^2$,we get

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$\left(\frac{10}{3}, \frac{11}{3}\right)$

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(D) Let $C = (x, y)$. Since $	riangle ABC$ is an isosceles right-angled triangle at $C$,we have $AC = BC$ and $AC \perp BC$.From $AC^2 = BC^2$,we get $(x-2)^2 + (y-3)^2 = (x-4)^2 + (y-5)^2$.Expanding this,$x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 10y + 25$.Simplifying,$4x + 4y = 28$,which gives $x + y = 7 \implies y = 7 - x$.Since $AC \perp BC$,the product of their slopes is $-1$: $\left(\frac{y-3}{x-2}ight) 	imes \left(\frac{y-5}{x-4}ight) = -1$.$(y-3)(y-5) = -(x-2)(x-4) \implies y^2 - 8y + 15 = -(x^2 - 6x + 8) \implies x^2 + y^2 - 6x - 8y + 23 = 0$.Substituting $y = 7-x$: $x^2 + (7-x)^2 - 6x - 8(7-x) + 23 = 0$.$x^2 + 49 - 14x + x^2 - 6x - 56 + 8x + 23 = 0 \implies 2x^2 - 12x + 16 = 0 \implies x^2 - 6x + 8 = 0$.$(x-2)(x-4) = 0$,so $x = 2$ or $x = 4$.If $x = 2$,$y = 5$,so $C = (2, 5)$. Centroid $G = \left(\frac{2+4+2}{3}, \frac{3+5+5}{3}ight) = \left(\frac{8}{3}, \frac{13}{3}ight)$.If $x = 4$,$y = 3$,so $C = (4, 3)$. Centroid $G = \left(\frac{2+4+4}{3}, \frac{3+5+3}{3}ight) = \left(\frac{10}{3}, \frac{11}{3}ight)$.Comparing with options,the correct answer is $\left(\frac{10}{3}, \frac{11}{3}ight)$.

Suppose $\triangle ABC$ is an isosceles triangle with $\angle C=90^{\circ}$,$A=(2,3)$ and $B=(4,5)$. Then the centroid of the triangle is

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