Finding coordinates of a point from bisector of two lines

I have two points, P1 and P2. Starting from M1 (middle point among P1 and the origin), I want to know the coordinates of the point Q1 as shown in the picture.

How can I do this?

Answer 1

[Star Strider]

Sure!

Fortunately, I kept the code ...

clear
close all

xq = [-1000 0 -200 -1000 -1000];
yq = [-450 0 -1000 -1000 -450];
a21 = atan2d(-450, -1000);
a23 = atan2d(-1000, -200);
a123 = a21 - a23;
ab = a123/2 + a23;
lineb = 1200*[cosd(ab); sind(ab)];
slopeb = lineb(2)/lineb(1);
len12 = hypot(-450, -1000);
xpt = interp1([0 -1000], [0 -450], -500);
ypt = interp1([0 -450], [0 -1000], xpt);
B = [0 1; lineb(2) 1] \ [0; lineb(1)];
slope_orth = -B(1);
intcpt_orth = ypt-slope_orth*xpt;
slope23 = -1000/-200;
xint = intcpt_orth/(slope23-slope_orth);
yint = slope_orth*xint+intcpt_orth;
figure
plot(xq, yq, DisplayName='Quadrilateral')
hold on
plot([0 lineb(1)], [0 lineb(2)],'-r', DisplayName='Bisector')
scatter(xint, yint, 25, 'red','filled', 's', DisplayName='Q1')
plot([0 -600], [0 -600]*slope_orth+intcpt_orth, '-m', DisplayName='Orthogonal Line')
hold off
text(xint+50, yint,"Q1 ("+xint+","+yint+")", Horiz='left', Vert='middle', FontWeight='bold')
axis('equal','padded')
legend(Location='NW')

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As a general rule, I delete my answers if they are not accepted.

.

[Gaetano Pavone]

Dear @Star Strider, can you please replace your answer?

[Gaetano Pavone]

@Star Striderthanks!

please, can you use the coordinates that I written in the question?

[Star Strider]

My pleasure!

I thought I did. I did not originally enlarge the image to see the exact coordinates.

This time, I did. Using them changes the result.

Try this --

clear
close all

xq = [-1000 0 -166 -1000 -1000];
yq = [-548 0 -1000 -1000 -548];
a21 = atan2d(-548, -1000);
a23 = atan2d(-1000, -166);
a123 = a21 - a23;
ab = a123/2 + a23;
lineb = 1200*[cosd(ab); sind(ab)];
slopeb = lineb(2)/lineb(1);
len12 = hypot(-548, -1000);
ypt = interp1([xq(2) xq(1)], [yq(2) yq(1)], -500);
xpt = interp1([0 yq(1)], [0 xq(1)], ypt);
B = [0 1; lineb(2) 1] \ [0; lineb(1)];
slope_orth = -B(1);
intcpt_orth = ypt-slope_orth*xpt;
slope23 = yq(3)/xq(3);
xint = intcpt_orth/(slope23-slope_orth);
yint = slope_orth*xint+intcpt_orth;
figure
plot(xq, yq, DisplayName='Quadrilateral')
hold on
plot([0 lineb(1)], [0 lineb(2)],'-r', DisplayName='Bisector')
scatter(xint, yint, 25, 'red','filled', 's', DisplayName='Q1')
plot([0 -600], [0 -600]*slope_orth+intcpt_orth, '-m', DisplayName='Orthogonal Line')
hold off
text(xq(1), yq(1), sprintf('P1 (%d, %d)',xq(1), yq(1)), Horiz='left', Vert='top')
text(xq(3), yq(3), sprintf('P2 (%d, %d)',xq(3), yq(3)), Horiz='left', Vert='top')
text(xpt, ypt, sprintf('M1 (%d, %d)',xpt,ypt), Horiz='left', Vert='bottom')
text(xint+50, yint,"Q1 ("+xint+","+yint+")", Horiz='left', Vert='middle', FontWeight='bold')
axis('equal','padded')
legend(Location='NW')
grid

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I added labels and coordinates for all the identified points. I also made subtle tweaks to my code, however the code itself, except for the changed coordinates, is the original. (I used degrees rather than radians here so that a visual check on the results would be easier to interpret.) I did not calculate the intersection of the bisector and the line connecting 'M1' and 'Q1'. I could add it if that information is desired.

.

[Gaetano Pavone]

@Star Strider it works very well! if you want to add further information feel free to update your code

[Star Strider]

Thank you.

I'll delete it again, since as I noted prevously, I usually delete my answers if they are not accepted.

It will remain posted if you accept it, not otherwise.

Answer 2

[Matt J]

M1=P1/2;

Q1=norm(M1)*P2/norm(P2);