Find the equatioon of straight line which are at a distance of 1 unit from the orgin and which passes through the point (3,1)?

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Answer 1 · 2017-08-10T11:27:36

The equation of line is # y=1 or 3x-4y -5=0#

The equation of line passing through #(3,1)# is #y-1 = m(x-3)#

Or # mx -y - 3m +1 =0 # . The distance of the origin#(0, 0)#

from the straight line is #1#. We know distance of a point

#(x_0,y_0)# from a straight line #ax+by+c=0# is

# d= (|ax_0+by_0+c|)/(sqrt(a^2+b^2)# , here

#a=m ; b=-1 , c=-3m+1 , d =1 , x_0=0 , y_0=0#

# :. 1= (|m*0 + (-1)*0+ (-3m+1)|)/(sqrt(m^2+1)) #

# (-3m+1) = sqrt(m^2+1) or (-3m+1)^2 = m^2+1# or

# 9m^2 -6m +1 =m^2+1 or 8m^2 -6m =0# or

#m(8m-6)=0 :. m=0 or m= 6/8=3/4#

The equation of line is when #m=0 ; y=1# and

the equation of line is when #m=3/4 ; y-1 = 3/4(x-3) #

#4y-4=3(x-3) or 3x -4y -5=0#

The equation of line is # y=1 or 3x-4y -5=0# [Ans]