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3. Some example problem based on section formula

1. Find the coordinates of the point which divides the line segment joining the points

$(3, -5)$ and

$(-3, 4)$ internally, in the ratio

$1 : 2$ .

Solution:

Let

$A(x_1, y_1) = (3, -5)$ ,

$B(x_2, y_2) = (-3, 4)$ and

$P(x, y)$ be the required point.

Ratio,

$(m : n) = 1 : 2$ .

Section formula:

$P (x, y) = (\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$

$= (\frac{1 \times (- 3) + 2 \times 3}{1 + 2}, \frac{1 \times 4 + 2 \times (- 5)}{1 + 2})$

$= (\frac{- 3 + 6}{3}, \frac{4 - 10}{3})$

$= (\frac{3}{3}, \frac{- 6}{3})$

$=$

$(1, -2)$

Therefore, the coordinates of the required point are

$(1, -2)$ .

2. Find the ratio in which the line segment joining the points

$(– 3, 10)$ and

$(6, – 8)$ is divided by

$(– 1, 6)$ .

Solution:

Let

$A(x_1, y_1) = (-3, 10)$ ,

$B(x_2, y_2) = (6, -8)$ and

$P(x, y) = (-1, 6)$ .

The ratio

$m : n$ can also be written as

$\frac{m}{n} : 1$ .

That is

$k : 1$ , where

$k = \frac{m}{n}$ .

If the point

$P(x, y)$ divides the line segment in the ratio

$k : 1$ , then the coordinates are:

$P (x, y) = (\frac{k x_{2} + x_{1}}{k + 1}, \frac{k y_{2} + y_{1}}{k + 1})$

$(- 1, 6) = (\frac{k \times 6 + (- 3)}{k + 1}, \frac{k \times (- 8) + 10}{k + 1})$

$(- 1, 6) = (\frac{6 k - 3}{k + 1}, \frac{- 8 k + 10}{k + 1})$

Equate the coordinate of

$x$ values.

$- 1 = \frac{6 k - 3}{k + 1}$

$-1(k + 1) = 6k - 3$

$-k - 1 = 6k - 3$

$-1 + 3 = 6k + k$

$2 = 7k$

$k = \frac{7}{2}$

So,

$k : 1 = \frac{7}{2} : 1 = 7 : 2$

Therefore, the ratio is

$7:2$ .

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3. Some example problem based on section formula

Theory: