Cartesian plane

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$If the vertices (x_{i}, y_{i}), i = 1, 2, 3 of the triangle lie on the curve x y = 1 and x_{i} satisfies the$
$equation x^{3} - 2 x^{2} + 7 x - 1 = 0, then the sum of coordinates of the centroid of the triangles is$

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Solution

$W e h a v e \frac{x_{1} + x_{2} + x_{3}}{3} = \frac{2}{3} A l s o y_{1} + y_{2} + y_{3} = \frac{1}{x_{1}} + \frac{1}{x_{2}} + \frac{1}{x_{3}} = \frac{x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1}}{x_{1} x_{2} x_{3}} = \frac{7}{1} = 7 \Rightarrow \frac{y_{1} + y_{2} + y_{3}}{3} = \frac{7}{3}$
$∴Sum of coordinates = \frac{2}{3} + \frac{7}{3} = 3$

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Cartesian plane

Remember concepts with our Masterclasses.

If the vertices xi,yi,i=1,2,3 of the triangle lie on the curve xy=1 and xi satisfies the equation x3−2x2+7x−1=0, then the sum of coordinates of the centroid of the triangles is

We have x1+x2+x33=23 Also y1+y2+y3=1x1+1x2+1x3=x1x2+x2x3+x3x1x1x2x3=71=7 ⇒y1+y2+y33=73∴Sum of coordinates =23+73=3

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$If the vertices (x_{i}, y_{i}), i = 1, 2, 3 of the triangle lie on the curve x y = 1 and x_{i} satisfies the$
$equation x^{3} - 2 x^{2} + 7 x - 1 = 0, then the sum of coordinates of the centroid of the triangles is$