Class 10: Coordinate Geometry

$\displaystyle \textbf{Question 1: }\text{Find the centroid of the triangle whose vertices are:}$
$\displaystyle \text{(i) }(1,4),\,(-1,-1),\,(3,-2)\qquad\text{(ii) }(-2,3),\,(2,-1),\,(4,0).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Centroid }=\left(\frac{1-1+3}{3},\frac{4-1-2}{3}\right)$
$\displaystyle =\left(1,\frac{1}{3}\right).$
$\displaystyle \therefore \text{The centroid is }\left(1,\frac{1}{3}\right).$

$\displaystyle \text{(ii) Centroid }=\left(\frac{-2+2+4}{3},\frac{3-1+0}{3}\right)$
$\displaystyle =\left(\frac{4}{3},\frac{2}{3}\right).$
$\displaystyle \therefore \text{The centroid is }\left(\frac{4}{3},\frac{2}{3}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Two vertices of a triangle are }(1,2),\,(3,5)\text{ and its centroid is at the origin.}$
$\displaystyle \text{Find the coordinates of the third vertex.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the coordinates of the third vertex be }(x,y).$
$\displaystyle \text{Since the centroid is }(0,0),$
$\displaystyle \left(\frac{1+3+x}{3},\frac{2+5+y}{3}\right)=(0,0).$
$\displaystyle \frac{4+x}{3}=0\quad\text{and}\quad\frac{7+y}{3}=0$
$\displaystyle x=-4,\quad y=-7.$
$\displaystyle \therefore \text{The coordinates of the third vertex are }(-4,-7).$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the third vertex of a triangle, if two of its vertices are at }(-3,1)\text{ and}$
$\displaystyle (0,-2)\text{ and the centroid is at the origin.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the coordinates of the third vertex be }(x,y).$
$\displaystyle \text{Since the centroid is }(0,0),$
$\displaystyle \left(\frac{-3+0+x}{3},\frac{1-2+y}{3}\right)=(0,0).$
$\displaystyle \frac{x-3}{3}=0\quad\text{and}\quad\frac{y-1}{3}=0$
$\displaystyle x=3,\quad y=1.$
$\displaystyle \therefore \text{The coordinates of the third vertex are }(3,1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }A(3,2)\text{ and }B(-2,1)\text{ are two vertices of a triangle }ABC\text{ whose centroid}$
$\displaystyle G\text{ has the coordinates }\left(\frac{5}{3},-\frac{1}{3}\right).\text{ Find the coordinates of the third vertex }C.$
$\displaystyle \hfill\text{[CBSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the coordinates of }C\text{ be }(x,y).$
$\displaystyle \text{Using the centroid formula,}$
$\displaystyle \left(\frac{3-2+x}{3},\frac{2+1+y}{3}\right)=\left(\frac{5}{3},-\frac{1}{3}\right).$
$\displaystyle \frac{1+x}{3}=\frac{5}{3}\quad\text{and}\quad\frac{3+y}{3}=-\frac{1}{3}$
$\displaystyle 1+x=5\quad\text{and}\quad3+y=-1$
$\displaystyle x=4,\quad y=-4.$
$\displaystyle \therefore \text{The coordinates of the third vertex }C\text{ are }(4,-4).$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{If }(-2,3),\,(4,-3)\text{ and }(4,5)\text{ are the mid-points of the sides of a triangle,}$
$\displaystyle \text{find the coordinates of its centroid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The centroid of the triangle formed by the mid-points is the same as that of the original triangle.}$
$\displaystyle \therefore \text{Centroid}=\left(\frac{-2+4+4}{3},\frac{3-3+5}{3}\right)$
$\displaystyle =\left(2,\frac{5}{3}\right).$
$\displaystyle \therefore \text{The coordinates of the centroid are }\left(2,\frac{5}{3}\right).$
$\displaystyle \\$