1.Find the distance between P(7, -4) and the line with equation x - 3y + 5 = 0 round to nearest tenth

Question

3Y=x+5
y=(x+5)/3
-1/m
= -3
y=-3x+b
sub in P(7,-4)
-4=-3(7)+b
b=21-4
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((7-4.6)^2+(-4-3.2)^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.

Answer 1

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)

Answer 2

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