A straight line passes through the point (-1,2) and it's distance from the origin is one unit. Find it's equation.

 

let x and y be the variables in the line. Then, by distance formula,
(x-0)2+(y-0)2=12.
x2+y2=1.
y2=1-x2.  (I) 
As per the standard equation for a line, 
y=mx+c
Squaring on both sides, we get:
y2=(mx+c)2 
   
=m2x2+2mxc+c2.  (II) 
From I and II,
1-x2 m2x2+2mxc+c2.
But we are given that (-1,2) is a point on this line. Therefore,
1-(-1)2 = m2(-1)2 + mc(2*-1) + c2.
1-1 = m2 - 2mc + c2.
(m-c)2 = 0
m-c = 0
m = c.  (III) 
Substituting in standard equation,
y = cx + c
y = (1+x)c.
Substituting the given point (-1,2), we get:
2 = (1-1)c




But here we are not getting c so recheck the question.

  • -25
let x and y be the variables in the line. Then, by distance formula,
(x-0)2+(y-0)2=12.
x2+y2=1.
y2=1-x2.  (I)
As per the standard equation for a line, 
y=mx+c
Squaring on both sides, we get:
y2=(mx+c)2
   =m2x2+2mxc+c2.  (II)
From I and II,
1-x2 = m2x2+2mxc+c2.
But we are given that (-1,2) is a point on this line. Therefore,
1-(-1)2 = m2(-1)2 + mc(2*-1) + c2.
1-1 = m2 - 2mc + c2.
(m-c)2 = 0
m-c = 0
m = c.  (III)
Substituting in standard equation,
y = cx + c
y = (1+x)c.
Substituting the given point (-1,2), we get:
2 = (1-1)c
??
 
  • -19
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