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Question

The distance of the point (1,2) from the line 3x+4y5=0 is units.

A
65
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B
56
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C
165
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Solution

The correct option is A 65
The distance of the point (x1,y1) from the line ax+by+c=0 is given by
|ax1+by1+c|a2+b2

Thus , the distance of the point (1,2) from the line 3x+4y5=0 is

d=|3×1+4×25|32+42

=|3+85|9+16

=|6|25=65 units.

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