how do you prove that (x1, y1), (x2,y2), (x3,y3) are collinear pts iff det of

lx1,y1,1l

lx2,y2,1l

lx3,y3,1l is 0.

the above is a 3x3 matrix.

Printable View

- November 23rd 2009, 12:27 PMalexandrabel90determinants
how do you prove that (x1, y1), (x2,y2), (x3,y3) are collinear pts iff det of

lx1,y1,1l

lx2,y2,1l

lx3,y3,1l is 0.

the above is a 3x3 matrix. - November 23rd 2009, 06:44 PMtonio
- November 23rd 2009, 10:35 PMalexandrabel90
sorry, i cant seem to see what you have typed:/

- November 23rd 2009, 11:27 PMShanks
The area of the triangle formed by the three given point is half of the absolute value of determinant.

Thus if the three points are colinear, then the determinant is equal to 0. - November 24th 2009, 12:41 AMalexandrabel90
what does it mean that the area formed by the 3 given pts is half the absolute value of the determinant?

could you explain further?

thank you! - November 24th 2009, 01:21 AMShanks
This is the formulla to find the area of a triangle formed by three points whose coordinates are given.