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The straight line passing through P(x1,y...

The straight line passing through `P(x_1,y_1)` and making an angle `alpha` with x-axis intersects `Ax + By + C=0` in Q then PQ=

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The correct Answer is:
`(|Ax_(1) + By_(1) + C|)/(|A "cos" alpha +B "sin" alpha|)`
Line is passing through the point P`(X_(1), Y_(1))` and making an angle `alpha` with x-axis.
One of the points on the line at distance r from the point P is given `(x_(1)+r "cos" alpha, y_(1) +r "sin" alpha)`.
If this point id Q, then it lies on the line. So, A`(x_(1) +r "cos" alpha) +B(y_(1) +r "sin" alpha) +C=0`.
`therefore r = (|Ax_(1) +By_(1)+C|)/(|A"cos" alpha + B"sin"alpha|)`
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