Coordinates of point X2Y2

[Guest]

Hi

Is it posssible to find coordinates X2,Y2 from the drawing? Give me a hint, please.

Data: X1Y1;X3Y3;X4Y4;L (L dimension always perpendicular to X1Y1-X4Y4); M (M dimension always perpendicular to X1Y1-X3Y3)
Find: X2Y2

Thanks
MK

[shyamjayakannan]

First, we find the equations of the lines passing through [tex](x_3,y_3),(x_1,y_1)[/tex] and [tex](x_4,y_4),(x_1,y_1)[/tex]:

[tex]\frac{y-y_1}{x-x_1}=\frac{y_3-y_1}{x_3-x_1}\Rightarrow y(x_3-x_1)-x(y_3-y_1)-y_1x_3+y_3x_1=0[/tex]
SImilarly, the other line is : [tex]y(x_4-x_1)-x(y_4-y_1)-y_1x_4+y_4x_1=0[/tex]

Use the formula for distance of a point from a line:
[tex]\frac{|y_2(x_3-x_1)-x_2(y_3-y_1)-y_1x_3+y_3x_1|}{\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}}=m[/tex] and [tex]\frac{|y_2(x_4-x_1)-x_2(y_4-y_1)-y_1x_4+y_4x_1|}{\sqrt{(x_4-x_1)^2+(y_4-y_1)^2}}=l[/tex]

Each equation will give you the equations for two lines(both cases for the modulus symbol) because there are two lines that are at a distance [tex]m[/tex] from the first line and similarly, for the second line. You need to decide which of the two you want to choose from the first equation. Then, choose the corresponding one from the second equation so that both the chosen lines are on the same side. Finally, [tex](x_2,y_2)[/tex] will be the point of intersection of the two lines so chosen.