Question

Find the third vertex of triangle ABC if two of its vertices are A (-2,3),B(4,5) and its centroid G(1,2)

Answer 1

Step-by-step explanation:

$( \frac{x1 + x2 + x3}{3} ).( \frac{y1 + y2 + y3}{3} ) = (x.y) \\ ( \frac{4 - 2 + x3}{3} ) = 1 \\ x3 = 3 \\ ( \frac{5 + 3 + y3}{3} ) = 2 \\ y3 = - 2 \\ therefore \: coordinates \: of \: third \: vertex \: are \\ (x3.y3) = (3. - 2)$

Answer 2

➢ Answer :

Coordinates of the third vertex (C) of the triangle = (1, -2)

➢ Given :

• Coordinates of vertex A = (-2, 3)
• Coordinates of vertex B = (4, 5)
• Coordinates of centroid G = (1, 2)

➢ To Find :

Coordinates of the third vertex (C).

➢ How To Find :

$\dagger$Let the coordinates of vertex A = $\bold{(x_1,y_1)}$

$\dagger$ Let the coordinates of vertex B = $\bold{(x_2,y_2)}$

$\dagger$ Let the coordinates of vertex C = $\bold{(x_3,y_3)}$

We know, formula for finding the centroid of a triangle =

$\bold{\left(\dfrac{x_1+x_2+x_3}{3}\right),\left(\dfrac{y_1+y_2+y_3}{3}\right)}$

Applying the formula for the given triangle, we get,

⇝ $\bold{\dfrac{-2+4+x_3}{3}=1}$

⇝ $\bold{\dfrac{x_3+2}{3}=1}$

⇝ $\bold{x_3+2=3}$

⇝ $\bold{x_3=1}$

Therefore, x coordinate of vertex C = 1.

$\hrulefill$

Again,

⇝ $\bold{\dfrac{3+5+y_3}{3}=2}$

⇝ $\bold{\dfrac{y_3+8}{3}=2}$

⇝ $\bold{y_3+8=6}$

⇝ $\bold{y_3=-2}$

Therefore, y coordinate of vertex C = -2

∴ Coordinates of the third vertex C =

$\bold{(x_3,y_3)=(1,-2)}$