Draw AL, PN and BM perpendiculars to the xy-plane.
Clearly the feet of the perpendiculars L, M, N are collinear and L = (x1,y1), M -(x2, y2), N = (x,y).
Now, draw a line through P which is parallel to the line LM meets AL produced at C and BM at D.
Clearly the triangle PAC and PDB are equiangular and hence similar.
![](https://www.sarthaks.com/?qa=blob&qa_blobid=10688760669295882609)
Since the lines AL, PN and BM are parallel, we have,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=3500319880549237816)
Remark:
(i) It P divides AB externally in the ratio m : n, then the coordinates of the point P are
![](https://www.sarthaks.com/?qa=blob&qa_blobid=7868632563002000742)
Observations:
- The point P divides AB internally in the ratio k : 1 iff k > 0 (i.e., k is + ve)
- The point P divides AB externally in the ratio |k| : 1 iff k < 0 (i.e., k is – ve)