How to calculate shortest distance from V (46,47) to straight line between points A (42,45) and B (52,40)?

Question

Must use Cartesian principles

Answer 1

Well, first you'd calculate the equation of the line AB, using the rise/run method. Then you find the slope of the perpendicular, which is -1/m if m represents the slope of AB. Then you calculate the point where the perpendicular, traveling from V to AB, crosses AB. That is the closest point on the line to the point.

Answer 2

First of all we have to locate the position of the point V. As per the data abscissa of the point V is 46 and of the initial end point of the line is 42 ie. 46 > 42. This implies that the point is within the ambit of the projection of the line.
Now, we have to understand whether the point is above the line or below. To know this we require slope of the line.
To find that consider the general equation y = mx + c
Then the slope of this line is, m = dy / dx
= {{y(2)-y(1)} / x(2) - x(1)}
Whence, x(1), x(2), y(1), y(2) are co-ordinates of the straight line AB.
Accordingly, the slope, m = (40-45)/(52-42) = - 0.5
Then c = 45 -(-0.5 * 42) = 66
Consequently, equation of the straight line is
y = -0.5 *x + 66
Now it is easy to find the shortest distance to the straight line from the point V. For that just enter the abscissa of the point in the equation of the straight line and get the corresponding ordinate of the straight line, finally find the difference from the ordinate of the point ; you get the shortest distance.
ie. y= (-0.5*46) + 66 = 43
Shortest distance is = 47 - 43 = 4
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Answer 3

first line A B y=mx+c
m=(52-42)/(40-45)=-.5
and at the point (42,45) ( any point on the line AB )
c=66
and the L ( the perpendicular between the point v and the line )
L=aG+bF+C\SQRT(A^2+B^2) (where g,f are 46,47 and a,b are -.5,-1 in the equation -.5x-y+66=0)
so L= 3.57
and i will be honest it been a while since i have taken that stuff so there is a 50 % chance that iam wrong

this answer

Answer 4

Imagine a point P on the line AB. Express it parametrically as a linear combination of the vectors A and B - let the parameter be t. Now write down expressions for vectors PB and PV. Write down the dot product PB.PV. This is an expression in terms of t. Find the value of t for which PB.PV = 0. Now you have both P and V, and pythagoras does the rest, or use a cross product (which amounts to the same thing). Or, to summarise, the answer is the cross product BPxBV, divided by the length of BP, for the P on the line AB for which PB.PV = 0.

(Sorry - had to edit this as I garbled it a bit earlier and mistyped some of the vectors, and forgot division by the length of BP. Incidentally, the problem gets much easier if you relocate the origin to (40,40), which doesn't affect the final answer)

Answer 5

first,we find the cartesian equation of AB

use the general formula

(y-y1)=(y2-y1)(x-x1)
/(x2-x1)...........(1)

letx1=42,y1=45,x2=52,y2=40

using(1), y-45=(40-45)(x-42)/(52-42)
= -(5/10)(x-42)

x+2y-132=0,..........(2)

the standard formula for the perpendicular
(shortist) distance of the point (x3,y3)from
the line ax+by+c=0 is
(ax3+by3+c/sqrt(a^2+b^2). .(3)

here,a= 1,b= 2,x3=46,y3=47,c= -132

using(3), ((1)(46)+(2)(47)-132)
/((1)^2+(2)^2)
=((46+94)-132)/sqrt 5 =8/(sqrt5)

therefore,the shortest distance from V to the
straight line between the points A and B is
(8/sqrt5)=3.577708764

JimA you got the same answer as i did

sqrt(1.6^2+3.2^2)
=sqrt((8/5)^2+(16/5)^2)
= sqrt(64/5) = 8/sqrt5

i hope that this helps

Answer 6

Line AB:
y=mx+c
m=delta y/delta x = (45-40)/(42-52)
y= - 0.5x+c
@ A
45= - 21 + c
=> c=66
y= -0.5x + 66

Normal to AB, passing through V
m = -1 / (-0.5)
=2
y=2x+c
@V
47=92+c
=> c=-45
y=2x-45

Intersection between normal and AB where
2x-45 = -0.5x+66
2.5x = 111
x = 111/2.5
= 44.4
y=2x-45
= 43.8

Distance between V and intersection
delta x = 46-44.4 = 1.6
delta y = 47 - 43.8 = 3.2

Distance = (1.6^2 + 3.2^2) ^ 0.5
= 2.58 (3sf)

You might want to double-check the numbers.

Answer 7

the shortest distance between any two points is zero

Answer 8

I suggest you should plot the points first before solving. This would help you understand more the problem and easily solve it.

Answer 9

sqrt (VA^2+(AB/2)^2)
or
sqrt (VB^2+(AB/2)^2)