If the vertices P, Q, R of a triangle PQR are | vedclass

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(A) Let the vertices be $P(x_1, y_1), Q(x_2, y_2),$ and $R(x_3, y_3),$ where $x_i, y_i \in \mathbb{Q}$.$1$. The Centroid is given by $\left( \frac{x_1

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Centroid

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(A) Let the vertices be $P(x_1, y_1), Q(x_2, y_2),$ and $R(x_3, y_3),$ where $x_i, y_i \in \mathbb{Q}$.$1$. The Centroid is given by $\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$. Since the sum and quotient of rational numbers are rational,the centroid is always a rational point.$2$. The Circumcentre $(x, y)$ satisfies the equations $(x-x_1)^2 + (y-y_1)^2 = (x-x_2)^2 + (y-y_2)^2$ and $(x-x_2)^2 + (y-y_2)^2 = (x-x_3)^2 + (y-y_3)^2$. These simplify to linear equations in $x$ and $y$ with rational coefficients. Thus,the circumcentre is always a rational point.$3$. The Incentre is given by $\left( \frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c} \right)$,where $a, b, c$ are side lengths. Since side lengths involve square roots (e.g.,$a = \sqrt{(x_2-x_3)^2 + (y_2-y_3)^2}$),they are generally irrational. Thus,the incentre is not necessarily a rational point.Therefore,both the Centroid and Circumcentre are always rational points. Given the options,the question implies identifying which of the listed points are rational. Since the Centroid and Circumcentre are rational,and the Incentre is not,the most appropriate choice is the one containing the rational points.

If the vertices $P, Q, R$ of a triangle $PQR$ are rational points,which of the following points of the triangle $PQR$ is (are) always a rational point $(s)$? ($A$ rational point is a point both of whose coordinates are rational numbers.)

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