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

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- May 22nd 2006, 02:39 AMhari-kjHow to draw an isosceles trianlge
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 - May 22nd 2006, 04:32 AMRich B.
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.

- May 22nd 2006, 05:16 AMhari-kj
a huge thanks for the solution.

but can you give me a definite formula for this say something like

x3=so and so

y3 =so and so

I'm sort of not able to do this.

Thanks again and sorry for bothering - May 22nd 2006, 02:41 PMThePerfectHacker
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. - May 22nd 2006, 06:14 PMThePerfectHackerQuote:

Originally Posted by**hari-kj**

- May 23rd 2006, 12:22 AMhari-kj
Thanks for the help guys.

I actually needed these data and this helped me alot.