find the coordinate of point on line x+y+3=0 whose distance from the line x+2y+2=0 is 5 unit 9
Answers
let the coordinates be (a, b)
As this point lies on line x + y + 3 = 0, the point (a, b) must satisfy this equation.
therefore we can write a + b = - 3 _______1
now it is given that the perpendicular distance from
point (a, b) to tne line x + 2y + 2 = 0 is √5 units,
| a + 2b + 2|/√(1² + 2²) = √5 [ it is √5 not 5 ]
[ perpendicular distance between any point (a1, b1) from a straight line ax + by + c = 0 is given by the equation |a1x + b1y + c|/√a1² + b1² ]
=> (a + 2b + 2)/√5 = √5
=> |a + 2b + 2| = √5 × √5
=> a + 2b + 2 = ± 5
=> a + 2b = 3 _________2
solving questions 1 and 2 [ taking + 5 ]
a + b = - 3
a + 2b = 3 (-)
_________
0 - b = - 6
=> b = 6
=> a + 6 = - 3
=> a = -9
therefore the coordinates are (6, - 9)
[ taking - 5 ]
a + 2b + 2 = -5
=> a + 2b = - 7
a + b = - 3 (-)
____________
0 + b = - 4
=> b = - 4
=> a + 2(-4) = - 7
=> a = - 7 + 8
=> a = 1
therefore other set of point is (-4, 1)