Question

Two of the vertices of a triangle are 5, -1 -3, -2 centroid is 1/3,5/3 find the coordinates of the third vertex third vertex also find area of triangle ABG by area of triangle ABC

Answer 1

The coordinate of the third vertex is (-1,8) and the area of the triangle is 39 sq. units.

• The two vertices of the triangle are given as (5,-1) and (-3,-2).
• Centroid of the triangle is given as (1/3,5/3)
• The formula to find the centroid of the triangle is given by

C(x,y) = ($\frac{x_{1}+x_{2}+x_{3} }{3}$,$\frac{y_{1}+y_{2}+y_{3} }{3}$)

where (x,y) are the points of the centroid and (x₁,y₁),(x₂,y₂) and (x₃,y₃) are the points of the triangle.

• Substituting the known values we solve for x- coordinate of the third vertex

$\frac{1}{3}$ = $\frac{5 - 3 + x_3}{3}$

x = -1

• Now calculating for y-coordinate of the third vertex of triangle

$\frac{5}{3}$ = $\frac{-1 - 2 + y_{3}}{3}$

y = 8

• The coordinates of the third vector is (-1,8).
• Now we have to calculate the area of triangle, we use the formula in coordinate geometry of area

Area = $|\frac{1}{2}[x_1(y_2 - y_3)+x_2(y_3 - y_1) + x_3(y_1 - y_2)] |$

• Substituting the values in the formula we get

Area = $|\frac{1}{2} [5(-2-8)+ (-3)(8-(-1)) + (-1)(-1 - (-2))]|$

Area = |$\frac{1}{2} [5(-10)+ (-3)(9) + (-1)(1)]$|

Area = |$\frac{1}{2} [-50-27-1]$|

Area = |(-78/2)|

Area = 39 sq. units.