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Question

Find the equation of the straight line which passes through the point (5,0) and is at a perpendicular distance of 4 units from origin.

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Solution

Let the equation of the line through points (5,0) be y0=m(x5)
or mxy5m=0(1)
The length of perpendicular from origin (0,0) on line is
m.005mm2+1=5mm2+1
it is given that 5mm2+1=4
25m2=16(m2+1)
9m2=16
m2=169 m=±43
So, when m=+43 from (1) equation of straight line
43xy5×43=04x3y20=0
where m=43 from (1) 43xy5(43)=04x+3y20=0

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