Math, asked by sanyabhatia29, 7 months ago

Find the coordinates of a point from the line x+y+3=0, whose distance from the line x+2y+2=0 is root75

Answers

Answered by amitnrw
1

Given : a point on  the line x+y+3=0, whose distance from the line x+2y+2=0 is root75

To Find : Coordinate of point

Solution:

Line x + 2y + 2  = 0

=> 2y = - x - 2

=> y = -(1/2)x - 1

slope = -1/2

Slope of perpendicular line

= 2

y = 2x  + c  

point lies on x + y + 3 = 0

=> 2x + c  =  -x - 3

=> 3x = -c - 3

=> x = -c/3  - 1

    y = 2x + c   =  2(-c/3 - 1) + c

=> y = c/3  - 2

(  -c/3  - 1 ,  c/3  - 2)

x+2y+2=0

y = 2x  + c  => 2y = 4x + 2c

x +  4x + 2c + 2 = 0

=> x = -(2/5)(c + 1)

y = 2 (-(2/5)(c + 1)) + c

=> y = c/5  -4/5

( -2c/5 - 2/5  , c/5 - 4/5)

(  -c/3  - 1 ,  c/3  - 2)  ( -2c/5 - 2/5  , c/5 - 4/5)

Distance = √75

( -c/3  - 1  + 2c/5 + 2/5)²  + (c/3  - 2 -  c/5 + 4/5)²  = 75

=> ( c/15 -3/5)²  + (2c/15 - 6/5)²   = 75

=> ( c/15 - 3/5)² = 15

=>  c/15 - 3/5 = ±√15

=> c - 9 = ±15√15

=>  c = 9  ±15√15

(  -c/3  - 1 ,  c/3  - 2)

= -3  -/+ 5√15 - 1  , 3  ±5√15 - 2

=  -4 -/+ 5√15 , 1  ±5√15

Points are  ( - 4 -  5√15 , 1 + 5√15)   and  ( - 4 +  5√15 , 1 - 5√15)

if distance was √5 instead of √75

( c/15 - 3/5)² = 1

=>  c/15 - 3/5 = ±1

=> c = 15( 3/5 ±1)

=> c =  9  ±15

=> c = 24  , -6

(  -c/3  - 1 ,  c/3  - 2)  

 (-9 , 6)  and (1 , -4)  are the points

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