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Question

Find the length of the perpendicular from the point (x1, y1) to the straight line Ax + By + C = 0, the axes being inclined at an angle ω, and the equation being written such that C is a negative quantity.


A

Ax1+By1+CA2+B2.cos ω

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B

Ax1+By1+CA2+B2.sin ω

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C

Ax1+By1+CA2+B22AB sin ω.cos ω

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D

Ax1+By1+CA2+B22AB cos ω.sin ω

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Solution

The correct option is D

Ax1+By1+CA2+B22AB cos ω.sin ω



Let the given straight line meet the axis in L and M. So that OL=CA and OM=CB
Let P be the given point (x1, y1). Draw the perpendicular PQ, PR and PS on the given line and two axes.
Taking O and P on opposite sides of the given line, we then have
ΔLPM+ΔMOL=ΔOLP+ΔOPM
i.e. PQ.LM + OL. OM sin ω = OL.PR + OM. PS - - - (1)
Draw PU and PV parallel to the axes of y and x, such that PU=y1 and PV = x1
Hence, PR = PU sin PUR =y1 sin ω
and PS = PV sin PVS = x1 sin ω
Also, LM=OL2+OM22OL.OM cos ω=C2A2+C2B22C2ABcos ω=C1A2+1B22 cos ωAB
Since, C is a negative quantity
On substituting these values in (1), we have
PQ×(C)×1A2+1B22 cos ωAB+C2ABsin ω=CAy1 sin ωCBx1 sin ω
On simplifying,
PQ=Ax1+By1+CA2+B22AB cos ω.sin ω


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