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Find the equations of two straight line passing through the point (0,1) on which the perpendiculars dropped from the point (2,2) are each of unit length.

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3 years ago

```							Let the equation of straight line be y=mx+c It passes through (0,a) a=m(0)+c⇒c=ay=mx+a........(i)mx−y+a=0 Lenght of perpendicular from (2a,2a) is equal to a 1+m 2 ​  ∣2am−2a+a∣​  =a∣2am−a∣=a 1+m 2 ​  4a 2 m 2 +a 2 −4a 2 m=a 2 +a 2 m 2 3a 2 m 2 −4a 2 m=03m 2 −4m=0m(3m−4)=0⇒m=0, 34​   Substituting m in (i) m=0 y=0(x)+ay=a m= 34​   y= 34​  x+a3y=4x+3a....(ii) Slope of line perpendicular to (i) is m ′ =− 43​   Equation of line passing (2a,2a) and slope m ′  is y−2a=− 43​  (x−2a)4y−8a=−3x+6a3x+4y=14a......(iii) Feet of perpendicular is point of intersection of (ii) and (iii) Solving (ii) and (iii), we get P( 56a​  , 513a​  ) Feet of perpendicular from (2a,2a) on y=a is Q(2a,a) Equation of PQ is  y−a= 56a​  −2a513a​  −a​  (x−2a)y−a=−2(x−2a)y+2x=5a Hence proved.
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4 months ago
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