- #1

- 3

- 0

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?

- Thread starter laura_jane
- Start date

- #1

- 3

- 0

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?

- #2

TD

Homework Helper

- 1,022

- 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.

[tex]\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[/tex]

- #3

- 3

- 0

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

- #4

TD

Homework Helper

- 1,022

- 0

- Replies
- 5

- Views
- 1K

- Replies
- 1

- Views
- 3K

- Replies
- 2

- Views
- 7K

- Replies
- 1

- Views
- 1K

- Replies
- 6

- Views
- 1K

- Replies
- 1

- Views
- 14K

- Replies
- 1

- Views
- 767

- Replies
- 8

- Views
- 7K

- Replies
- 6

- Views
- 714

- Replies
- 8

- Views
- 1K