A line through point origin meets the line 2x= 3y +13 at right angles at point Q .find the cdinates of Coordinates of Q?

1 Answer
Mar 14, 2018

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The equation of the given straight line is

#2x= 3y +13#

#=>2x-3y =13#

#=>2/sqrt(2^2+3^2)x-3/sqrt(2^2+3^2)y =13/sqrt(2^2+3^2)#

#=>2/sqrt13x-3/sqrt13y =13/sqrt13#

#=>xcosalpha+ysinalpha =p#

Where #2/sqrt13=cosalpha;-3/sqrt13=sinalpha ;13/sqrt(2^2+3^2)=sqrt13=p#
This is the normal form of the given equation and #p# is the length of the perpendicular from origin.

The coordinates of foot #(Q)# of the perpendicular from origin to the straight line will be

#(p cosalpha,p sinalpha)#

#=(sqrt13xx2/sqrt13,sqrt13xx(-3/sqrt13 ))#

#=(2,-3)#