Posted by Jon on Saturday, April 25, 2009 at 3:43pm.
1.Find the distance between P(7, 4) and the line with equation x  3y + 5 = 0 round to nearest tenth
3Y=x+5
y=(x+5)/3
1/m
= 3
y=3x+b
sub in P(7,4)
4=3(7)+b
b=214
b=17
y=3x+17
set the two equations equal.
3x+17=(x+5)/3
9x+51=x+5
46=10x
x=4.6
sub back in:
y= 3(4.6)+17= 13.8+17=3.2
distance=sqrt((74.6)^2+(43.2)^2)
=sqrt((2.4)^2+(7.2)^2)
=sqrt(5.76+51.84)
=sqrt(57.6)=7.59
=7.6
Is there a shorter way to do this?
2.Write XY(with line over it) as the sum of unit vectors for X(8,2,9) and Y(12,1,10).
There's an example just like this in my book but I really don't understand it.

Math  Count Iblis, Saturday, April 25, 2009 at 6:27pm
There is indeed a shorter way. What you do is you shift the origin of the coordinate system so that the line moves to the origin.
To do that just find a random point that lies on the line, say, the point
(5,0). Then if we translate the entire coordinate system so that this point moves to the orgin, the point P will have coordinates
(7, 4)  (5,0) = (12, 4)
Now consider the unit vector e1 that points in the direction along the line and the unit vector e2 that points orthogonal to the line.
If you express the point P = (12,4) in terms of e1 and e2, like:
P = r e1 + s e2
then you can interpret this as moving from the origin to P as moving along the line over a distance r and orthogonal to the line over a distance s.
So, clearly all we need to do is expand P in terms of e1 and e2 and then the coefficient of e2 is the answer.
e2 is, of course, proportional to
(1,3). You have to normalize it:
e2 = 1/sqrt(10) (1,3)
An then s follows from the general expansion formula of vectors in terms of unit vectors:
P = (P dot e1) e1 + (P dot e2) e2
s = P dot e2 =
(12,4) dot (1,3)/sqrt(10) =
24/sqrt(10)

Math  Reiny, Sunday, April 26, 2009 at 11:49am
Or an even shorter way is to use the formula for the distance from a point (p,q) to the line Ax + By + C = 0
= │Ap + Bq + C│/√(A^2 + B^2)
= │7 +12 +5│/(1+9)
= 24/√10
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