Cartesian plane

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The image of centroid of the triangle formed by the points $(0, 0), (1, 2), (2, 1)$ in y- axis is $(h, k)$ then $h + k$ is

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Solution

The centroid of the triangle is G
$\begin{matrix} G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}) \\ = (\frac{0 + 1 + 2}{3}, \frac{0 + 2 + 1}{3}) \\ = (1, 1) \end{matrix}$
The image of point $(a, b)$ in y- axis is $(- a, b)$
Therefore, the image of $(1, 1)$ in y- axis is $(h, k) = (- 1, 1)$
It implies that $h + k = 0$

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Cartesian plane

Remember concepts with our Masterclasses.

The image of centroid of the triangle formed by the points 0,0,1,2,2,1 in y- axis is h,k then h+k is

The centroid of the triangle is GG=x1+x2+x33,y1+y2+y33=0+1+23,0+2+13=1,1The image of point a,b in y- axis is −a,bTherefore, the image of 1,1 in y- axis is h,k=−1,1It implies that h+k=0

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The image of centroid of the triangle formed by the points $(0, 0), (1, 2), (2, 1)$ in y- axis is $(h, k)$ then $h + k$ is