Question

find the coordinates of the point of the intersection of the medians of the triangle ABC; given A(-2,-3), B(6,7) and C(4,1)

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Answer 1

Answer

Coordinates of centriod is 8/3 and 5/3

$\rule{100}2$

Explanation

Given:-

• ABC is a triangle.
• Given that A(-2, -3), B(6, 7) and C(4, 1)

Find:-

Coordinates of the point if intersection of the medians of the triangle ABC.

Solution:-

We know that -

Centroid : Intersection of medians of triangle.

We can find coordinates of centroid, using formula -

For x-axis

$\Rightarrow\:\sf\bigg( \dfrac{ x_1 \: + \: x_2\: + \: x_3 }{3} \bigg)$

For y-axis

$\Rightarrow\:\sf\bigg( \dfrac{ y_1 \: + \: y_2\: + \: y_3 }{3} \bigg)$

We have -

• $\sf{x_1}$ = -2
• $\sf{x_2}$ = 6
• $\sf{x_3}$ = 4

• $\sf{y_1}$ = -3
• $\sf{y_2}$ = 7
• $\sf{y_3}$ = 1

Put these values in above formula

First for x-axis

$\implies\:\sf\bigg( \dfrac{ -2\: + \: 6\: + \: 4 }{3} \bigg)$

$\implies\:\sf\bigg( \dfrac{ -2\: + \: 10 }{3} \bigg)$

$\implies\:\sf\bigg( \dfrac{ 8 }{3} \bigg)$

For y-axis

$\implies\:\sf\bigg( \dfrac{ -3\: + \: 7\: + \: 1 }{3} \bigg)$

$\implies\:\sf\bigg( \dfrac{ -3\: + \: 8 }{3} \bigg)$

$\implies\:\sf\bigg( \dfrac{ 5 }{3} \bigg)$

•°• Coordinates of the centroid is $\sf\bigg( \dfrac{8}{3}, \:\dfrac{ 5 }{3} \bigg)$

Answer 2

Question : Find the coordinates of the point of the intersection of the medians of the triangle ABC; given A(-2, -3), B(6, 7)and C(4, 1).

______________________________

Solution : Here, we have $(x_1= -2,\:y_1=- 3),(x_2=6,\:y_2 = 7)</p><p>and\\(x_3=4,\:y_3=1)$

$\implies$ Let P(x, y) be the centroid of the triangle ABC then,

$\implies$ x = $\frac{x_1+x_2+x_3}{3}=\frac{(-2) +6+4}{3}= \frac{-2+10}{3}=\frac{8}{3}$

$\implies$ y = $\frac{y_1+y_2+y_3}{3}=\frac{(-3) +7+1}{3}= \frac{-3+8}{3}=\frac{5}{3}$

$\implies$ $\therefore$ The coordinates of the centroid P of the triangle ABC are $(\frac{8}{3}, \frac{5}{3})$

Thus, the coordinates of the point of intersection of the medians of triangle is are $(\frac{8}{3}, \frac{5}{3})$.