 QUESTION

# Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of $\Delta ABC$.(i). The median from A meets BC at D. Find the coordinates of the point D.(ii). Find the coordinates of the point P on AD such that AP: PD = 2: 1(iii). Find the coordinates of points Q and R on medians BE and CF respectively such that BQ: QE =2:1(iv). What do you observe? [Note: The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1] (v). If A $({x_1},{y_1})$, B $({x_2},{y_2})$ , and C $({x_3},{y_3})$ are the vertices of $\Delta ABC$, find the coordinates of the centroid of the triangle.

Hint: The section formula for a line segment joined by the points $({x_1},{y_1})$ and $({x_2},{y_2})$ which is divided by the point (x, y) in the ratio m:n is given by $(x,y) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$. Use this formula to find the answers.

The points are given as A (4, 2), B (6, 5) and C (1, 4).

The section formula for a line segment joined by the points $({x_1},{y_1})$ and $({x_2},{y_2})$ which is divided by the point (x, y) in the ratio m:n is given by:

$(x,y) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$

D is the midpoint of the line segment joining the points B and C. The coordinates of the point D is given by:

$D = \left( {\dfrac{{6 + 1}}{2},\dfrac{{5 + 4}}{2}} \right)$

$D = \left( {\dfrac{7}{2},\dfrac{9}{2}} \right)$

The coordinates of point P, that divides the line segment AD in the ratio 2:1 is given as follows:

$P = \left( {\dfrac{{2\left( {\dfrac{7}{2}} \right) + 4}}{3},\dfrac{{2\left( {\dfrac{9}{2}} \right) + 2}}{3}} \right)$

$P = \left( {\dfrac{{7 + 4}}{3},\dfrac{{9 + 2}}{3}} \right)$

$P = \left( {\dfrac{{11}}{3},\dfrac{{11}}{3}} \right)$

The midpoint E of line AC is given as follows:

$E = \left( {\dfrac{{4 + 1}}{2},\dfrac{{2 + 4}}{2}} \right)$

$E = \left( {\dfrac{5}{2},3} \right)$

The coordinates of point Q, that divides the line segment BE in the ratio 2:1 is given as follows:

$Q = \left( {\dfrac{{2\left( {\dfrac{5}{2}} \right) + 6}}{3},\dfrac{{2\left( 3 \right) + 5}}{3}} \right)$

$Q = \left( {\dfrac{{5 + 6}}{3},\dfrac{{6 + 5}}{3}} \right)$

$Q = \left( {\dfrac{{11}}{3},\dfrac{{11}}{3}} \right)$

The midpoint F of line segment AB is given as follows:

$F = \left( {\dfrac{{4 + 6}}{2},\dfrac{{2 + 5}}{2}} \right)$

$F = \left( {5,\dfrac{7}{2}} \right)$

The coordinates of point R, that divides the line segment CF in the ratio 2:1 is given as follows:

$R = \left( {\dfrac{{2\left( 5 \right) + 1}}{3},\dfrac{{2\left( {\dfrac{7}{2}} \right) + 4}}{3}} \right)$

$R = \left( {\dfrac{{10 + 1}}{3},\dfrac{{7 + 4}}{3}} \right)$

$R = \left( {\dfrac{{11}}{3},\dfrac{{11}}{3}} \right)$

We observe that all the medians intersect at a common point $\left( {\dfrac{{11}}{3},\dfrac{{11}}{3}} \right)$, which divides the median in the ratio 2:1. This common point is called the centroid of the triangle.

If A $({x_1},{y_1})$, B $({x_2},{y_2})$ and C $({x_3},{y_3})$, then the midpoint D of BC is given as:

$D = \left( {\dfrac{{{x_2} + {x_3}}}{2},\dfrac{{{y_2} + {y_3}}}{2}} \right)$

The coordinates of point P, that divides the line segment AD in the ratio 2:1 is given as follows:

$P = \left( {\dfrac{{2\left( {\dfrac{{{x_2} + {x_3}}}{2}} \right) + {x_1}}}{3},\dfrac{{2\left( {\dfrac{{{y_2} + {y_3}}}{2}} \right) + {y_1}}}{3}} \right)$

$P = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$

The coordinates of the centroid of the triangle is $\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$.

Note: Even though you know the formula and definition of the centroid, proceed with the solution as stated in the question, otherwise, you may not get full credits for your answer.