Question

If the points (2a,a),(a,2a) and (a,a) enclose a triangle of area 18 square units. Then the centroid of the triangle is:

Answer 1

Answer:

Centroid is (8,8) or (-8,-8)

Step-by-step explanation:

Formula used:

Area of the triangle formed by the points

$(x_1, y_1),\:(x_2, y_2)\:and \:(x_3, y_3)\:is\:\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Centroid of a triangle having vertices

$(x_1,y_1),\:(x_2,y_2)\:and\:(x_3,y_3)\: is\:(\frac{x_1+x_2+x_3 }{3},\frac{y_1+y_2+y_3 }{3})$

Given points are (2a, a), (a, 2a), (a,a)

Area of the triangle = 18 square units

$\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=18$

$\frac{1}{2}[2a(2a-a)+a(a-a)+a(a-2a)]=18$

$2a(2a-a)+a(a-a)+a(a-2a)=18*2$

$2a(a)+a(0)+a(-a)=36$

$2a^2+0-a^2=36$

$a^2=36$

$a=6, -6$

Centroid of the given triangle

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

$=(\frac{2a+a+a}{3},\frac{a+2a+a}{3})$

$=(\frac{4a}{3},\frac{4a}{3})$

when a=6,

the centroid is $(\frac{4(6)}{3},\frac{4(6)}{3})$

$(8,8)$

when a=-6,

the centroid is $(\frac{4(-6)}{3},\frac{4(-6)}{3})$

$(-8,-8)$