Question

the coordinates of centroid of the triangle are 1,3 and two of its vertices are-7,6 and8,5. find the third vertex.also,find the coordinates of the centroid of the triangle with the third vertex 2,4

Answer 1

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Answer 2

Answer:

$</p><p>Third \: vertex =(x,y)=( 2,-5)$

$Centroid = (1,5)$

Step-by-step explanation:

The centroid of a triangle is the point of intersection of its medians .

$The \: coordinates \: of \: \\the \: centroid \: are \: \big(\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3}\big)$

$i)Given ,\\Centroid (G)=(1,3)$

$and, \: (-7,6),(8,5)\:are \: two \: verticies \: of \: a \: triangle .$

$Let ,\: third \: vertex = (x,y)$

$\big(\frac{-7+8+x}{3},\frac{6+8+y}{3}\big)=(1,3)$

$\big(\frac{1+x}{3},\frac{14+y}{3}\big)=(1,3)$

$\frac{1+x}{3}=1\: ,\frac{14+y}{3}=3$

$1+x=3\: ,14+y=9$

$x=3-1\: ,y=9-14$

$x = 2 \: ,y = -5$

$Therefore,\\</p><p>Third \: vertex =(x,y)=( 2,-5)$

$ii)(-7,6),(8,5) \: and \: (2,4)\: are \: \\Three\: verticies\: of \: a \: triangle$

$Centroid =\big(\frac{-7+8+2}{3},\frac{6+5+4}{3}\big)\\=\big(\frac{3}{3},\frac{15}{3}\big)\\=(1,5)$

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