Coordinate Geometry

[Dionysus22]

Quick question, have tried and failed on quite a few occasions:



K is the line 3x - 4y + 9 = 0

The point a(-3, 0) is on K

b(x, y) is a point on K such that | ab | = 15 and x > 0.

Find the value of x and the value of y.
(Ans (9, 9))



I've tried using the distance of a line formula but I end up with too many variables.
Any help is greatly appreciated.

[HallsofIvy]

Quick question, have tried and failed on quite a few occasions:



K is the line 3x - 4y + 9 = 0

The point a(-3, 0) is on K

b(x, y) is a point on K such that | ab | = 15 and x > 0.

Find the value of x and the value of y.
(Ans (9, 9))



I've tried using the distance of a line formula but I end up with too many variables.
Any help is greatly appreciated.
|ab| is the length of the segment. The distance from any point (x,y) to (-3, 0) is sqrt((x+3)^3+ y^2) so you are looking for (x,y) that satisfy sqrt((x+3)^2+ y^2)= 15 or (x+3)^2+ y^2= 225. That's one equation in two unknowns but you also know (x,y) must satisfy 3x - 4y + 9 = 0 in order to be on the given line. I would be inclined to solve the second equation for x:
x= (4/3)y- 3 and put that into the first equation: x+3= (4/3)y- 3+ 3= (4/3)y so (x+3)^2+ y^2= 225 becomes ((4/3)y)^2+ y^2= 225 or (16/9)y^2+ y^2= (25/9)y^2= 225. y^2= 225(9/25)= 81 so y= 9 or -9. Putting those back into 3x- 4y+ 9= 0, with y= 9 we get 3x- 36+9= 0 or 3x= 27 so x= 9. With y= -9, 3x+ 36+ 9= 0 so 3x= -45 so x= -15.

Since you were told to find the point with x> 0, (x,y)= (9, 9).

[Dionysus22]

Thanks very much,

had tried that way but obviously my basic maths let me down.