Finding a Perpendicular Point on a Triangle with Given Coordinates

[laura_jane]

if A(1,2) and B(8,3) find any point P on the x-axis such that AP is perpendicular to BP.

Here's what I know:
P (P,0)
perpendicular slopes are opposite reciprocals of each other
ABP is a right triangle

any ideas how to start this problem?

 

[TD]

We have A(1,2), B(8,3) and P(p,0).

Then vector PA = A-P = (1-p,2) and vector PB = B - P = (8-p,3).

You want those vectors to be perpendicular so their inner product has to be 0.

\left( {1 - p,2} \right) \cdot \left( {8 - p,3} \right) = 0 \Leftrightarrow \left( {1 - p} \right)\left( {8 - p} \right) + 6 = 0 \Leftrightarrow p = 7 \,\,\vee \,\, p = 2

 

[laura_jane]

Thanks very much, I forgot that perpendicular slopes multiplied to zero. You helped a ton!

 

[TD]

No problem, don't forget we're talking about the scalar (or inner) product of vectors though, not just multiplication.