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
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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