Name: Distance Between Two Points
Uploaded: 2019-07-19
Duration: 4 min 36 s
Description: We calculate the area of a circle using a formula. Similarly, we can calculate the distance between two points using the distance formula. Watch this video t...

Question

Let

A (4, 2), B (6, 5)

and

C (1, 4)

be the vertices of

△ ABC

. Justify that the point which divides each median in the ratio

2 : 1

is the centroid of a triangle such that

AP : PD = 2 : 1

.

Accepted Answer

Here, we have

From the figure, we observe that the point $P$ divides the side $A D$ in a ratio $2 : 1$ .
Since $D$ is the midpoint of $C (1, 4)$ .
So, $D = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{1}}{2}) = (\frac{6 + 1}{2}, \frac{5 + 4}{2}) = (\frac{7}{2}, \frac{9}{2})$

According to the section formula, we know that $P (x, y) = (\frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}})$
$\Rightarrow P (x, y) = (\frac{2 \times \frac{7}{2} + 1 \times 4}{2 + 1}, \frac{2 \times \frac{9}{2} + 1 \times 2}{2 + 1})$
$\Rightarrow P (x, y) = (\frac{11}{3}, \frac{11}{3})$
Thus, the coordinates of $P$ are $(\frac{11}{3}, \frac{11}{3})$ .

Let us say $O$ and $N$ divides the sides $BE$ and $CF$ respectively in the ratio $2 : 1$ .
Where $E$ and $F$ are the midpoints of $AC$ and $AB$ respectively.

Coordinates of $E = (\frac{4 + 1}{2}, \frac{2 + 4}{2}) = (\frac{5}{2}, 3)$
And coordinates of $F = (\frac{6 + 4}{2}, \frac{5 + 2}{2}) = (5, \frac{7}{2})$

Thus, we have the points $A (4, 2), B (6, 5), C (1, 4), E (\frac{5}{2}, 3), F (5, \frac{7}{2})$ .

According to the section formula, $O$ divides $BE$ in the ratio $2 : 1$ , we know that $O (x, y) = (\frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}})$ $\Rightarrow O (x, y) = (\frac{2 \times \frac{5}{2} + 1 \times 6}{2 + 1}, \frac{2 \times 3 + 1 \times 5}{2 + 1})$
$\Rightarrow O (x, y) = (\frac{11}{3}, \frac{11}{3})$

Thus, the coordinates of $O$ are $(\frac{11}{3}, \frac{11}{3})$

Again by section formula, $N$ divides $CF$ in the ratio $2 : 1$ , we have

$N (x, y) = (\frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}})$ $\Rightarrow N (x, y) = (\frac{2 \times 5 + 1 \times 1}{2 + 1}, \frac{2 \times \frac{7}{2} + 1 \times 4}{2 + 1})$

$\Rightarrow N (x, y) = (\frac{11}{3}, \frac{11}{3})$

Thus, the points $P (\frac{11}{3}, \frac{11}{3}), O (\frac{11}{3}, \frac{11}{3}), N (\frac{11}{3}, \frac{11}{3})$ coincides each other.

This coinciding point is called the centroid of the triangle.

Hence justified.

Let $A (4, 2), B (6, 5)$ and $C (1, 4)$ be the vertices of $△ ABC$ . Justify that the point which divides each median in the ratio $2 : 1$ is the centroid of a triangle such that $AP : PD = 2 : 1$ .

Important Questions on Coordinate Geometry

Learn and Prepare for any exam you want !

Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of △ABC. Justify that the point which divides each median in the ratio 2:1 is the centroid of a triangle such that AP:PD=2:1.

Important Questions on Coordinate Geometry

Learn and Prepare for any exam you want !

Let $A (4, 2), B (6, 5)$ and $C (1, 4)$ be the vertices of $△ ABC$ . Justify that the point which divides each median in the ratio $2 : 1$ is the centroid of a triangle such that $AP : PD = 2 : 1$ .