MATLAB Answers

How can I plot a triangle with its altitudes?

조회 수: 8(최근 30일)
Jack Bason
Jack Bason 30 Apr 2018
편집: John BG 11 May 2018
Hi I'd just like to know whats wrong with my coding i have as when i run this it does not plot the altitudes correctly.
x = [0,2,4,0];
y = [0,3,2,0];
%Altitude CF
% Gradient Forumla
% m = (y2-y1)/(x2-x1)
m3 = (y(2)-y(1))/(x(2)-x(1));
altitudem3=-1/m3;
% c = y-(m*x)
base3_c = y(1)-(m3*x(1));
altitude3_c = y(3)-(altitudem3*x(3));
% y=mx+c
% mx+base_c=altitudemx+altitude_c
% x=((altitude_c)-(base_c))/(m-altitudem)
xintersection=(altitude3_c-base3_c)/(m3-altitudem3);
yintersection=(altitudem3*xintersection)+altitude3_c;
%Altitude BE
m2 = (y(3)-y(1))/(x(3)-x(1));
altitudem2=-1/m2;
% c = y-(m*x)
base2_c = y(1)-(m2*x(1));
altitude2_c = y(2)-(altitudem2*x(2));
% y=mx+c
% mx+base_c=altitudemx+altitude_c
% x=((altitude_c)-(base_c))/(m-altitudem)
xintersection2=(altitude2_c-base2_c)/(m2-altitudem2);
yintersection2=(altitudem2*xintersection)+altitude2_c;
%Altitude AD
m = (y(3)-y(2))/(x(3)-x(2));
altitudem=-1/m;
% c = y-(m*x)
base_c = y(2)-(m*x(2));
altitude_c = y(1)-(altitudem*x(1));
% y=mx+c
% mx+base_c=altitudemx+altitude_c
% x=((altitude_c)-(base_c))/(m-altitudem)
xintersection3=(altitude_c-base_c)/(m-altitudem);
yintersection3=(altitudem*xintersection)+altitude_c;
hold on
plot(x,y);
plot([xintersection,x(3)],[yintersection,y(3)])
plot([xintersection2,x(2)],[yintersection2,y(2)])
plot([xintersection3,x(1)],[yintersection3,y(1)])
hold off

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Jan
Jan 3 May 2018
The "altitude" is the line through a vertex, which is orthogonal to the opposite edge.
John BG
John BG 4 May 2018
Thanks Jan
.
I already supplied an answer satisfying the request of the question, the core point to solve being the calculation of such 'altutides' or normal segments to each triangle side.
The question originator added the request that in my opinion should be a new question to show the intersects of the altitudes, besides the intersects of the altitudes on each triangle side.
The answer you have supplied doesn't catch all altitudes for the following particular cases:
1. missing 1 altitude
x = [0,0,4]
y = [0,3,2]
.
.
2. missing 2 altitudes
x = [0,0,4]
y = [0,4,0]
.
.
the horizontal and vertical sides of the triangle should be coloured with the altitudes, but they aren't because this answer doesn't get these altitudes
.
My answer catches such altitudes missed by Jan's answer
.
x = [0,0,4]
y = [0,3,2]
.
x = [0,0,4]
y = [0,4,0]
.
.
John BG
John BG
John BG 5 May 2018
Also, none of the supplied answers so far, including mine, consider negative coordinates of the input points
x =[ -6 -5 -4] y =[ -10 4 10]
.

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채택된 답변

Jan
Jan 4 May 2018
편집: Jan 4 May 2018
If the question is: "I'd just like to know whats wrong with my coding", this is the answer:
x = [0,2,4,0];
y = [0,3,2,0];
% This is fine:
m3 = (y(2)-y(1))/(x(2)-x(1));
altitudem3=-1/m3;
base3_c = y(1)-(m3*x(1));
altitude3_c = y(3)-(altitudem3*x(3));
xintersection=(altitude3_c-base3_c)/(m3-altitudem3);
yintersection=(altitudem3*xintersection)+altitude3_c;
m2 = (y(3)-y(1))/(x(3)-x(1));
altitudem2=-1/m2;
base2_c = y(1)-(m2*x(1));
altitude2_c = y(2)-(altitudem2*x(2));
xintersection2=(altitude2_c-base2_c)/(m2-altitudem2);
% yintersection2=(altitudem2*xintersection)+altitude2_c;
% ^ Here is the problem
% It must be xintersection2:
yintersection2=(altitudem2*xintersection2)+altitude2_c;
m = (y(3)-y(2))/(x(3)-x(2));
altitudem=-1/m;
base_c = y(2)-(m*x(2));
altitude_c = y(1)-(altitudem*x(1));
xintersection3=(altitude_c-base_c)/(m-altitudem);
yintersection3=(altitudem*xintersection)+altitude_c;
% ^ Here is the problem
% It must be xintersection3:
yintersection3=(altitudem*xintersection3)+altitude_c;
hold on
plot(x,y);
plot([xintersection,x(3)],[yintersection,y(3)])
plot([xintersection2,x(2)],[yintersection2,y(2)])
plot([xintersection3,x(1)],[yintersection3,y(1)])
This is a typical problem, when code is duplicated by copy&paste. Modifying all occurrences of the differences is tricky. A more reliable solution is to use a loop instead. See John BG's answer, which considers m==0 in addition. But a code as near as possible to your code:
x = [0,2,4,0];
y = [0,3,2,0];
hold on
plot(x,y);
v = mod(1:5, 3) + 1;
for k = 1:3
i1 = v(k);
i2 = v(k+1);
i3 = v(k+2);
m = (y(i2)-y(i1)) / (x(i2)-x(i1));
altitudem = -1/m;
base_c = y(i1) - m * x(i1);
altitude_c = y(i3) - altitudem * x(i3);
xintersection(k) = (altitude_c - base_c) / (m - altitudem);
yintersection(k) = (altitudem * xintersection(k)) + altitude_c;
plot([xintersection(k), x(i3)], [yintersection(k), y(i3)])
end
Using loops instead of copy&past + modifications reduces the chance for typos.

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추가 답변(2개)

sloppydisk
sloppydisk 30 Apr 2018
Please clarify what you are trying to do. What would you like to see as output?

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Jack Bason
Jack Bason 30 Apr 2018
As an output I would like to see a plotted triangle with all 3 altitudes that intersect at one point. The coding needs to have the ability to change the co-ordinates and it still work.

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John BG
John BG 1 May 2018
편집: John BG 1 May 2018
Hi Jack Bason
I have rearranged your code to do what you asked for, please find attached copy of this script.
close all;clear all;clc
x=[0,2,4];y=[0,3,2];
x2=[x x x(1)];y2=[y y y(1)];
plot(x2,y2,'b')
hold on
x_intersect=zeros(1,3)
y_intersect=zeros(1,3)
for k=1:1:length(x)
p1=[x2(k) y2(k)]
p2=[x2(k+1) y2(k+1)]
pref=[x2(k+2) y2(k+2)]
% 1st segment slope: m offset: k0
m1=(p2(2)-p1(2))/(p2(1)-p1(1))
switch abs(m1)
case Inf % vertical, 90.00 degree up
yint=pref(2)
xint=p1(1) % = p2(1), xint
m2=0
case 0 % horizontal, completely flat, 0.00 degree
yint=p1(2) % =p2(2)
xint=pref(1) % = p2(1)
m2=0
otherwise
k01=p1(2)-m1*p1(1)
m2=-1/m1 % 2nd segment m k0
k02=pref(2)-m2*pref(1)
xint=(k01-k02)/(m2-m1) % intersect point
yint=m1*xint+k01
x_intersect(k)=xint % logging intersect point
yIntersect(k)=yint
end
plot([p1(1) p2(1)],[p1(2) p2(2)],'g*')
plot(pref(1),pref(2),'y*')
plot(xint,yint,'g*')
plot([xint pref(1)],[yint pref(2)],'r')
end
.
.
if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?
To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link
thanks in advance for time and attention
John BG

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