Question

The centroid of a triangle is (2, 3) and two of
it's vertices are (5, 6) and (-3,4). The third
vertex of the triangle is:​

Answer 1

Given,

• A = (5,6)

• B = (-3,4)

• C = (x,y)

• O = (2,3)

Vertex of centroid ca be determined by,

$\boxed{\sf{Centroid=(\frac{x_1+x_2+x_3}{3}),(\frac{y_1+y_2+y_3}{3})}}}$

Applying the values,

$\implies\sf{2=\frac{5-3+x}{3}}\:\:\:\:,\:\:\:\:\implies 3=\frac{6+4-y}{3}$

$\implies\sf{2=\frac{2+x}{3}}\:\:\:\:,\:\:\:\:\implies 3=\frac{10-y}{3}$

$\implies\sf{6=2+x}\:\:\:\:,\:\:\:\:\implies 9=10-y$

$\implies\sf{x=6-2}\:\:\:\:,\:\:\:\:\implies y=10-9$

$\implies\sf{x=4}\:\:\:\:,\:\:\:\:\implies y=1$

$\boxed{\sf{Third \:vertex =(4,1)}}$

Answer 2

Provided:-

• Centroid of the triangle = (2,3)
• Two of the vertices of triangle = (5,6) and (-3,4)

To FinD:-

• The third vertex of the triangle?

How to solve?

Here, we have been given Centroid of the triangle which is the intersecting point of three meridians of the triangle.

• The centroid of the triangle is given by $\sf{( \frac{x_1 + x_2 + x_3}{3} ), (\frac{y_1 + y_2 + y_3}{3}})$ where (x1, y1), (x2, y2) and (x3, y3) are the vertices of the triangle.

So, By using this formula, Let's solve the Q.

Solution:-

We have,

• Vertices of the triangle (5,6) and (-3,4) and centroid of the triangle (2,3)

Let,

• Assume the third vertex be (x3, y3)

$\large{ \sf{ \longrightarrow{(x \: y) = (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} )}}}$

$\large{ \sf{ \longrightarrow \: (2 \: 3) = ( \frac{5 + ( - 3) + x_3}{3} , \frac{6 + 4 + y_3}{3} )}}$

By comparing both sides,

|| Evaluating x3 and y3 ||

➝ 5 + (-3) + x3 /3 = 2

➝ 2 + x3 = 6

➝ x3 = 4

✒ x coordinate of third vertex = 4

&

➝ 6 + 4 + y3 /3 = 3

➝ 10 + y3 = 9

➝ y3 = -1

✒ y coordinate of third vertex = -1

☀️ So the coordinates of the third vertex of the triangle is (4, -1) [Answer]

━━━━━━━━━━━━━━━━━━━━