If i have two points (x1,y1) and (x2,y2)
What formula would I use to find the third point so that it forms an isosceles triangle.
Guys please help me out.
I'm stuck up in this problem.
Regards,
Bala K J
Hi:
Picture, if you will, the segment with end points P(x1,y1) and Q(x2,y2). Now picture a line perpendicular to, and passing through the segment's midpoint. Any point on this line will serve as the third point, i.e., the triangle's vertex. [true because i) the vertex of an isosceles triangle is equidistant to the endpoints of the base, and ii) A given point is equidistant to two other points, P1 and P2, iff it lies on the perpendicular bisector of segment (P1P2). So, it stands to reason that the coordinates the third point satisfy the equation of said perpendicular line.
To that end, two lines are perpendicular iff their slopes are negative reciprocals of one another. Thus, m=-(x2-x1)/(y2-y1) =(x2-x1)/(y1-y2). From m=(y-yk)/(x-xk) for some point (xk,yk) on the line, y=m(x-xk)+yk. The only point we know of is the midpoint of the segment with coordinates (0.5(x1+x2), 0.5(y1+y2)).
Conclusion: The third point is any point whose coordinates satisfy equation y=m(x-xk)+yk, where xk=0.5(x1+x2), yk=0.5(y1+y2), and m=(x2-x1)/(y1-y2).
Regards,
Rich B.
You have two points,
$\displaystyle P_1(x_1,y_1)$ and $\displaystyle P_2(x_2,y_2)$
Find a point(s) $\displaystyle P(x,y)$ such as you form an isoseles triangle. Meaning the distance $\displaystyle PP_1=PP_2$.
Using distance formula,
$\displaystyle \sqrt{(x-x_0)^2+(y-y_0)^2}=\sqrt{(x-x_1)^2+(y-y_1)^2}$
Square,
$\displaystyle (x-x_0)^2+(y-y_0)^2=(x-x_1)^2+(y-y_1)^2$
Regroup,
$\displaystyle (x-x_0)^2-(x-x_1)^2=(y-y_1)^2-(y-y_0)^2$
Use difference of two squares,
$\displaystyle (2x-(x_1+x_0))(x_1+x_0)=(2y-(y_0+y_1))(y_0+y_1)$
So all the points are a line that satisfy this equation.