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Section Formula

The section f...

Question

The section formula (internal division only) says that if point $R (X, Y)$ divides the join of $P (X, Y)$ and $Q (x_{2}, y_{2})$ internally in a given ratio $(m_{1} : m_{2})$ , then coordinates of point R are

A

(\begin{matrix} \frac{m_{1} y_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, \frac{m_{1} x_{1} + m_{2} y_{2}}{m_{1} + m_{2}} \end{matrix})

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B

(\begin{matrix} \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}, \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}} \end{matrix})

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C

(\begin{matrix} \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}} \end{matrix})

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D

None of the above

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Solution

The correct option is C $(\begin{matrix} \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \end{matrix})$
Section formula for internal division:

$x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}}$

So, option C is correct.

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Q. The coordinates of the point P(x, y) which divides the line segment joining the points

A (x_{1}, y_{1})

B (x_{2}, y_{2})

internally in the ratio

m_{1} : m_{2}

(\frac{m_{1} x_{2} - m_{2} x_{1}}{m_{1} + m_{2} 1}, \frac{m_{1} y_{2} - m_{2} y_{1}}{m_{1} + m_{2}})

Q. If a point P(x, y) divides the line segment joining , A

(x_{1}, y_{1})

(x_{2}, y_{2})

m_{1}

m_{2}

then the coordinates of P are

(p \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}, q \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}})

R (X, Y)

is mid point of the line joining

P (X, Y)

Q (x_{2}, y_{2})

m_{1} = m_{2}

, then the coordinates of R are

y + b = m_{1} (x + a), y + b = m_{2} (x + a)

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