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find distance of the point(1,2) from the line 4x + 7y +18= 0 along the line x-y=8.
Asked by | 05 Mar, 2013, 01:06: PM Expert Answer
equation of line pependicular to the 4x+7y+18=0
and passing through (1,2)
7x-4y=-1
perpendicular distance of (1,2) from the line 4x+7Y=8
36/√65
angle between x-y=8 and 7x-4y=-1
Φ=tan^-1[(7/4-1)/1+7/4]
=tan^-1[3/11]
distance of (1,2) from the line 4x+7y+18=0 along x-y=8
=rcosΦ=36/√65
r=36/√65*11/√130
=36*11/65√2
Answered by | 06 Mar, 2013, 02:02: PM
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