Question

A straight line has non -zero intercepts a and b on the X axis and Y axis. find the centroid of the triangle formed with the axes and the origin

Answer 1

$\textbf{Concept:}$

$\text{Centroid of a triangle having vertices }$ $(x_1,y_1),\;(x_2,y_2)$ $\text{ and }$ $(x_3,y_3)$$\text{ is }$

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

$\text{since the line make the intercepts a and b,}$

$\text{it cut the x axis at (a,0) and the y axis at (0,b)}$

$\text{Now, }$

$\text{Centroid of the triangle formed by the points (a,0), (0,b) and (0,0) is}$

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

$(\frac{a+0+0}{3},\;\frac{0+b+0}{3})$

$\bf(\frac{a}{3},\;\frac{b}{3})$