If A (5, –1), B(–3, –2) and C(–1, 8) are the vertices of triangle ABC, find the length of median through A and the coordinates of the centroid.

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Solution

Let AD be the median through the vertex A of $△ A B C$ . Then, D is the mid-point of BC. So, the coordinates are $(\frac{- 3 - 1}{2}, \frac{- 2 + 8}{2})$ i.e., (-2, 3)
Distance between the points is given by
$\sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}}$

$∴ A D = \sqrt{(5 + 2)^{2} + (- 1 - 3)^{2}}$

= $\sqrt{49 + 16} = \sqrt{65}$ units

Let G be the centroid of $△$ ABC. Then, G lies on median AD and divides it in ratio 2:1. So, coordinates of G are using section formula
$(x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}})$

$(\frac{2 \times (- 2) + 1 \times 5}{3}, \frac{2 \times 3 + \times (- 1)}{2 + 1})$

= $(\frac{- 4 + 5}{3}, \frac{6 - 1}{3})$ = $\frac{1}{3}, \frac{5}{3}$

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[4 MARKS]

△ A B C

is a triangle with vertices

A = (x, x + 3), B = (2, 1)

and

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5

Coordinates of the centroid of

△ A B C

are

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, the coordinates of vertices

A

B

and

C

are

(4, 7)

(- 2, 3)

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respectively. Find the equations of the medians passing through the vertices

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