Multivariable Calculus
Solution to Problem 49, Page 821
Use a scalar projection to show that the distance from a point P1 (x1 , y1 ) to the line ax + by + c = 0 is
|ax1 +by1 +c|
√
. Use this formula to find the distance between the point (−2, 3) to the line 3x − 4y + 5 = 0.
a2 +b2
Solution:
Let Q(x0 , y0 ) be any point on the line L = {(x, y) : ax + by + c = 0}. The distance we seek is the length of
−−→
the projection of vector QP 1 onto any vector that is orthogonal to L. We claim that a vector orthogonal to
L is the red-black vector ~v = ha, bi, illustrated below. The red part depicts the desired length.
y
−
−
→
QP 1
P1 (x1 , y1 )
Q(x0 , y0 )
−−→
QR
R(x2 , y2 )
~v = ha, bi
x
L = {(x, y) : ax + by = c}
To see why this claim is true suppose R(x2 , y2 ) is another point on L. Then the direction of L can be
−−→
represented by the vector QR = hx2 −x0 , y2 −y0 i. The claim that ha, bi is orthogonal to L can be established
−−→
by showing that QR • ha, bi = 0. The relevant calculations follow:
−−→
QR • ha, bi = hx2 − x0 , y2 − y0 i • ha, bi
= (ax2 + by2 ) − (ax0 + by0 )
= −c − (−c)
(because both (x0 , y0 ) and (x2 , y2 ) are on L)
= 0.
−−→
−−→
Now, QP 1 = hx1 − x0 , y1 − y0 i, and the length of the projection of QP 1 onto ha, bi is (see page 835)
−−→
|hx1 − x0 , y1 − y0 i • ha, bi|
|QP 1 • ha, bi|
√
=
kha, bik
a2 + b2
|ax1 − ax0 + by1 − by0 |
√
=
a2 + b2
|ax1 + by1 − (ax0 + by0 )|
√
=
.
a2 + b2
The point Q(x0 , y0 ) is on L, so ax0 + by0 + c = 0. Thus, Expression (1) reduces to
|ax1 + by1 + c|
√
,
a2 + b2
which is what we wanted to prove.
Applying this formula to the point (−2, 3) and the line 3x − 4y + 5 = 0 gives
distance =
|3(−2) − 4(3) + 5|
13
√
=
.
5
9 + 16
(1)
Follow-up Question:
Why do we put absolute values in the numerator of this distance formula (the formula in the book does
not have them)? Sure, so we get a positive number, but why is this procedure legitimate, and what does a
negative number indicate?
Extra credit to all legitimate explanations submitted by the beginning of class on Wednesday, September 16.
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