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How can I plot a triangle with its altitudes?

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Jack Bason
Jack Bason on 30 Apr 2018
Edited: John BG on 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
  4 Comments
John BG
John BG on 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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Accepted Answer

Jan
Jan on 4 May 2018
Edited: Jan on 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.

More Answers (2)

sloppydisk
sloppydisk on 30 Apr 2018
Please clarify what you are trying to do. What would you like to see as output?
  1 Comment
Jack Bason
Jack Bason on 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 on 1 May 2018
Edited: John BG on 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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