Question

A straight line has non zero intercepts a and b an the X axis and Y axis Find the centroid of the triangle formed with the axes and the origin​

Answer 1

THE COORDINATES OF CENTROID IS GIVEN BY = $(\frac{a}{3} \frac{b}{3})$

• A straight line has non zero intercepts a and b on x and y axis respectively . this means the two vertices of triangle are A(a,0)  and B(0,b) and the third vertices of the triangle is origin that is C(0,0).

• We know that the coordinates of centroid is given by

• $X_{1}=\frac{a_{1} + b_{1} +c_{1} }{3} \\Y_{1}=\frac{a_{2} +b_{2}+ c_{2}}{3}$

• $X_{1}=\frac{a+0+0}{3}$=$\frac{a}{3}$

• $Y_{1}=\frac{0+b+0}{3} = \frac{b}{3}$

• hence coordinates of centroid are $(\frac{a}{3} \frac{b}{3})$

thanks!!1

Answer 2

Coordinate Geometry

Intercept form of a straight line when intercepts on the axes made by the straight line are given:

• Let a line segment intersects the axes at the distances $a$ and $b$ respectively. Then the equation of the straight line be
• $\quad\quad\frac{x}{a}+\frac{y}{b}=1$
• Here the above straight line intersects the $x-$ axis at $(a,\:0)$ and the $y-$ axis at $(0,\:b)$.

Formula to find centroid of a triangle:

• Let the vertices of a triangle $\Delta ABC$ be $A\:(x_{1},\:y_{1})$, $B\:(x_{2},\:y_{2})$ and $C\:(x_{3},\:y_{3})$.
• Then the coordinates of its centroid be
• $\quad\quad \big(\frac{x_{1}+x_{2}+x_{3}}{3},\:\frac{y_{1}+y_{2}+y_{3}}{3}\big)$

Solution:

• Given that, the straight line has non-zero intercepts $a$ and $b$ on the $x-$ axis and $y-$ axis respectively.
• Then the straight line intersects the coordinate axes at $A\:(a,\:0)$ and $B\:(0,\:b)$ respectively.
• Also given that, a triangle $\Delta ABO$, whose vertices are $A\:(a,\:0)$, $(0,\:b)$ and the origin $O\:(0,\:0)$.
• $\therefore$ the centroid of the triangle is at
• $\quad\quad \big(\frac{a+0+0}{3},\:\frac{0+b+0}{3}\big)$
• $\quad\Rightarrow \big(\frac{a}{3},\:\frac{b}{3}\big)$

Answer:

The centroid of the triangle formed with the axes and the origin is at $\big(\frac{a}{3},\:\frac{b}{3}\big)$.