Question

In this question you will derive a general formula for the distance of a point from a line. Let P be the point (p,q) and L the line y=mx+b.
The slope of L is
.

The slope of a line perpendicular to L is
.

The line through P perpendicular to L can be written as y=sx+c
where s is:

and c is:
.
That line intersects L in the point Q=(u,v),

where u is:

and v is:
.

The distance of P and Q is
.

(Note, all answers must be in terms of m, b, p and q.)

Answer 1

@one2345

Answer 2

the slope of L is m

Answer 3

just kidding i didn't get all of it

Answer 4

The slope of a line perpendicular to L is -1/m?

Answer 5

the slope of a line that is perpindicular is -1/m

Answer 6

yes

Answer 7

The line through P perpendicular to L can be written as y=sx+c
where s is: ?

and c is: ?

Answer 8

@jayzdd

Answer 9

s = -1/m

Answer 10

c= ?

Answer 11

@jayzdd

Answer 12

@jayzdd
That line intersects L in the point Q=(u,v),

where u is: ?

and v is: ?

The distance of P and Q is ?

Answer 13

A) m
B) -1/m
C)
line through (p,q) with slope -1/m
y - q = -1/m ( x - p)
y = -1/m * x + p/m + q
s = -1/m
c = p/m + q

Answer 14

whats u and v? @jayzdd

Answer 15

That line intersects L in the point Q=(u,v),

where u is:

and v is:
.

The distance of P and Q is
.

Answer 16

its easier if we graph this

Answer 17

so what does that mean

Answer 18

whats u, v & The distance of P and Q is

Answer 19

that is what you are given, a point P(x,y) and a line L given by the equation y = mx + b. We want to find the distance from P(x,y) to line L

Answer 20

ok

Answer 21

to do this, we draw a perpendicular line through P(x,y)
the equation of this perpendicular line is
y - q = -1/m ( x - p)
solving for y
y = -1/m * x + p/m + q

Answer 22

ok

Answer 23

ok

Answer 24

what is the correct answer for x?

Answer 25

Now we want to find the intersection of the perp. line and the line L .
We will call this intersection point Q(u,v).
At the intersection point the y values of the two lines are the same
y= mx + b
y = -1/m*x + p/m + q

set them equal
mx + b = -1/m*x + p/m + q
solve for x

mx +1/m* x = p/m + q - b

x ( 1 + 1/m) = p/m + q -b

x = ( p/m + q -b) / (1 + 1/m)

Answer 26

and i still need

That line intersects L in the point Q=(u,v),

where u is:

and v is:

Answer 27

ok

Answer 28

whats u & v

Answer 29

That line intersects L in the point Q=(u,v),

where u is:

and v is:

Answer 30

u = ( mq + p - bm) / (1 + m^2)

Answer 31

whats v?

Answer 32

@jayzdd

Answer 33

v = (m^2*q+m*p+b)/(m^2+1)

Answer 34

u and distance were wrong

Answer 35

i didnt find the distance yet

Answer 36

show me a screenshot of the error

Answer 37

oh what is it

Answer 38

it just says its not correct for u

Answer 39

u said u is u = ( mq + p - bm) / (1 + m^2)

Answer 40

got u

Answer 41

need distance and then we're done :)

Answer 42

$$u = \frac{-bm + p + mq} {m^2 + 1}$$

Answer 43

got u need distance

Answer 44

its correct now?

Answer 45

i put it in wrong u had it correct

Answer 46

i just need the distance now