Equation of line in Straight lines

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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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Equation of line in Straight lines

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=1,1 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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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