find the equation of the straight line passing through the point ( 3, -4 ) and perpendicular to the line 5x + 3y - 1 = 0
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Given Eqn. --> 5x + 3y - 1 = 0
==> y = 1/3[1 - 5x] = -5/3 x + 1/3
--> Slope = -5/3...
---> Slope of perpendicular line = -1/ ( -5/3) = 3/5
==> The eqn. of line is --> [ y - (-4) ] = 3/5 [ x - 3 ]
==> Req.d Eqn. = 5y + 20 = 3x - 9
==> Req.d Eqn. = 3x - 5y - 29 = 0 <-- Ans..
Given Eqn. --> 5x + 3y - 1 = 0
==> y = 1/3[1 - 5x] = -5/3 x + 1/3
--> Slope = -5/3...
---> Slope of perpendicular line = -1/ ( -5/3) = 3/5
==> The eqn. of line is --> [ y - (-4) ] = 3/5 [ x - 3 ]
==> Req.d Eqn. = 5y + 20 = 3x - 9
==> Req.d Eqn. = 3x - 5y - 29 = 0 <-- Ans..
pradeepkumar7:
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