Silly.girl said:
Alright well i've been having a lot of trouble trying to figure out how to find the distance between a line and a point not on the line.
So if anyone could help me i'd really like that.
Here are some sample problems:
Sample 1:
(-1,4) and y=x-1
Sample 2:
(-3,1) and y+4=1/3<--one-third(x-2)
if you could take them step by step for me i'd really like that.
thanks.
The following formula may be used to find the distance between a point (x<SUB>1</SUB>, y<SUB>1</SUB>) and the line whose equation is
Ax + By + C = 0:
d = Ax<SUB>1</SUB> + By<SUB>1</SUB> + C
sss--------------------
ssss ± √(A<SUP>2</SUP> + B<SUP>2</SUP>)
The sign of the radical in the denominator is chosen to be OPPOSITE the sign of C.
I'll help you with the second problem, because it is a bit more involved.
The equation of the line is y + 4 = (1/3)(x - 2)
The first step is to get the equation of the line into the form
Ax + By + C = 0
so that we can identify the values of A, B, and C.
Multiply both sides of the equation by 3:
3*y + 3*4 = 3*(1/3)*(x - 2)
3y + 12 = 1(x - 2)
3y + 12 = x - 2
Next, -x + 2 to both sides of the equation:
-x + 2 + 3y + 12 = x - 2 - x + 2
-x + 3y + 14 = 0
Let's multiply both sides by -1 to get the coefficient of x to be positive:
x - 3y - 14 = 0
Now, we can see that A = 1, B = (-3), and C = -14.
And we want the distance between this line and the point (-3, 1).
Just substitute into the formula:
d = 1(-3) - 3(1) - 14
sss------------------
ssss + √ [(1)<SUP>2</SUP> + (-3)<SUP>2</SUP>]
Note that because C is negative, we use the + value of the radical in the denominator. Now, it is just arithmetic:
d = -3 -3 - 14
sss---------------
ssss √10
or, d = -20 / √10
Multiply numerator and denominator by √10 to rationalize the denominator, and reduce the fraction:
d = -2 √10
The fact that the distance turned out to be negative tells you that the point and the origin are on the SAME side of the line; if the distance turned out to be positive, you'd know that the point and the origin are on OPPOSITE sides of the line.
In most application situations, we are concerned just with the absolute value of d; in this case | d | = 2 √10
