Question

Find the equation of straight line passing through the centroid of the triangle formed the vertices (3,-4),(-2,1) and (5,0) and parallel to X-3Y=4.

Answer 1

Answer:

I'm solving this problem graphically..



Represent 2x+ 3y-6=0 graphically first.

Triangle formed is tri ABC, Vertices A(0,0). B(3,0) & C(0,2)

AB = 3 unit

=> mid point M is ( 1.5,0)

In the above shown triangle ABC, centroid is represented by poit ‘P'. Which is the point of concurrency of 3 medians.

Since, centroid divides each median in the ratio 2:1

=> CP:PM = 2:1

Hence, coordinates of centroid P(x,y) can be calculated by section formula.

x = (m1x2 + m2x1)/(m1 + m2)

& y = (m1y2+m2y1)/(m1+m2)

here, m1= 2, m2=1, x1= 0, x2= 1.5, y1= 2, y2= 0

x = (2*1.5 + 1*0)/(2+1)

=> x = 3/3 = 1 . . . . . . . .(1)

y = ( 2*0 + 1*2) / (2+1)

=> y = 2/3 . . . . . . .(2)

=> centroid is P ( 1, 2/3)