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Distance between Parallel Lines

Find the rati...

Question

Find the ratio in which the line ax + by + c = 0 divides the line segment joining p $(x_{1}, y_{1})$ and Q $(x_{2}, y_{2})$ .

You are given t = $\frac{(a x_{2} + b y_{2} + c)}{a x_{1} + b y_{1} + c}$

-t

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Solution

The correct option is C

The ratio is $\frac{m}{n} = \frac{(a x_{1} + b y_{1} + c)}{a x_{2} + b y_{2} + c}$ ,which is $\frac{- 1}{t}$ .

We can derive this formula in the following way.

First assume the ratio is K:1. Let the point of intersection be R. We can find the co-ordinates of R interms of K. We know R lies on the other line also. Substitute the co-ordinates in the given line to find the value of K.

We can use this approach while solving problems. It is good to remember the formula as well.

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Similar questions

Show that the plane $a x + b y + c z + d = 0$ divides the line joining the points

$(x_{1}, y_{1}, z_{1}) a n d (x^{2}, y^{2}, z^{2})$ in the ratio

$- \frac{a x_{1} + b y_{1} + c z_{1} + d}{a x_{2} + b y_{2} + c z_{2} + d}$ .

P; a x + b y + c z + d = 0

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

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

(- \frac{{a x}_{1} + {b y}_{1} + {c z}_{1} + d}{{a x}_{2} + {b y}_{2} + {c z}_{2} + d}) = - \frac{p_{1}}{p_{2}}

P_{1} = {a x}_{1} + {b y}_{1} + {c z}_{1} + d

P_{2} = {a x}_{2} + {b y}_{2} + {c z}_{2} + d

L = a x + b y + c = 0

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

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

\frac{- L_{1}}{L_{2}}

L_{i} = a x_{i} + b y_{i} + c

(x_{1}, y_{1})

m : n

\frac{n x}{x_{1}} + \frac{m y}{y_{1}} = m + n

- \frac{a x_{1} + b y_{1} + c z_{1} + d}{a x_{2} + b y_{2} + c z_{2} + d}

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