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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: