intersection point

[pmss]

Hello

Some one can help me to find the intersection point between two lines. One
line is irregular and another is straight. For example I have given here two
lines here . I have to find out the intersection point between these two
lines. Thank YOu

First Line
x y
0 80
5 105
15 110
25 120
35 160
50 150
Second Line
x y
0 60
25 140

Pmss

 

[Joel]

Using both lines as linear the answer is x = 1.35, y = 88.7

I added both lines into a chart and added a trendline to both graphs to get
the following equations

y=80x-20
y=15x + 68.33. this line it was difficult to dtermine if I shoud use a
linear approximation or some other approximation.

then I solved manually for x and y

80x-20 = 15x + 68.33
65x = 88.33
x = 1.3589

y= 80x - 20
y = 80 * (1.3589) - 20
y = 1.35

 

[pmss]

Thank you.
But i have not only these two lines , everytime x and y is changing
according to my data. Means i have many lines. So is it possible to find out
intersection point using VB code ? If yes please.

Thank you
pmss

 

[Jon Peltier]

Andy Pope has done this:

http://andypope.info/charts/intersection.htm

- Jon

 

[Dana DeLouis]

I added both lines into a chart and added a trendline to both graphs to

get
the following equations
y=80x-20

For the data set
x y
0 60
25 140

Hi. Just some feedback. When given data pairs like above, the linear
equation should have a y-intercept of 60 by inspection.
(The slope (80/25 or 3.2)

I believe what you did was use a "Line" graph in which the x-points are
numbered 1,2...etc
This would give an equation of 80*x-20

A Scatter Chart would have placed the x-points in the correct location, and
given an equation of:

3.2 x + 60

Again, just some thoughts.

If line-segment interception is being used, then I show the two lines as:
60.+ 3.2 x
95 + x

Hence:
x: 15.909090909090908
y: 110.9090909090909