Intersection of line and curve from thier points
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I have a line with 2 points on it and a curve with 4 points on it
For the line we have :
x_line = [7020 0]
y_line = [0 250000]
For the curve we have :
x_curve = [5620 4850 4290 3600]
y_curve = [0 100000 200000 300000]
I need to know the intersection point of the line and the curve
Any help will be appreciated.
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Star Strider
2017년 12월 8일
Try this:
x_line = [7020 0];
y_line = [0 250000];
x_curve = [5620 4850 4290 3600];
y_curve = [0 100000 200000 300000];
b_line = polyfit(x_line, y_line,1); % Fit ‘line’
y_line2 = polyval(b_line, x_curve); % Evaluate ‘line’ At ‘x_curve’
x_int = interp1((y_line2-y_curve), x_curve, 0); % X-Value At Intercept
y_int = polyval(b_line,x_int); % Y-Value At Intercept
figure(1)
plot(x_line, y_line)
hold on
plot(x_curve, y_curve)
plot(x_int, y_int, '+r') % Plot Intercept Point
hold off
댓글 수: 1
Juan Sebastián Gómez Chitiva
2021년 1월 24일
exactly what i need!!! Thanks!!!!
this is the best answer i've seen
추가 답변 (3개)
Mohamed Ameen
2017년 12월 8일
댓글 수: 1
Star Strider
2017년 12월 8일
As always, my pleasure.
First, ‘line’ is a linear function, so creating an equation for it (using polyfit) is straightforward. I cannot compare the y-values for ‘y_line’ and ‘y_curve’ directly, so I evaluate ‘line’ at the ‘x_curve’ points to create y-values at those points as ‘y_line2’. I then subtract ‘y_curve’ from them, and use interp1 to find the x-value (as ‘x_int’) where they are 0. I then use polyval with ‘x_int’ to get the y-value at the interception, ‘y_int’.
The plots are then straightforward.
Jamie Hetherington
2019년 12월 24일
Starstrider this solution was excellent. I needed a method to identify the point of intersection between a straight line and a plotted curve of residuals. I was thinking it could be done solving simulatenous equations but would've taken more time to create and implement.
Great stuff!
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