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Grade: 11

                        

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.

3 years ago

Answers : (1)

Arun
24735 Points
							
Let the equation of straight line be y=mx+c
 
It passes through (0,a)
 
a=m(0)+c
⇒c=a
y=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=0
3m 
2
 −4m=0
m(3m−4)=0
⇒m=0, 
3
4
​ 
 
 
Substituting m in (i)
 
m=0
 
y=0(x)+a
y=a
 
m= 
3
4
​ 
 
 
y= 
3
4
​ 
 x+a
3y=4x+3a....(ii)
 
Slope of line perpendicular to (i) is
 
 =− 
4
3
​ 
 
 
Equation of line passing (2a,2a) and slope m 
  is
 
y−2a=− 
4
3
​ 
 (x−2a)
4y−8a=−3x+6a
3x+4y=14a......(iii)
 
Feet of perpendicular is point of intersection of (ii) and (iii)
 
Solving (ii) and (iii), we get
 
P( 
5
6a
​ 
 , 
5
13a
​ 
 )
 
Feet of perpendicular from (2a,2a) on y=a is
 
Q(2a,a)
 
Equation of PQ is 
 
y−a= 
5
6a
​ 
 −2a
5
13a
​ 
 −a
​ 
 (x−2a)
y−a=−2(x−2a)
y+2x=5a
 
Hence proved.
4 months ago
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