MHB Distance between points question

[Yankel]

Hello

I have two points: B(1,3) and C(2,6). I need to find a point A on the y-axis, from which BC is "seen" at an angle of 90 degrees.

I tried using Pythagoras theorem and got:

\[y^{2}-6y+10=y^{2}-12y+40+10\]

but it doesn't give the correct answer, which is (0,5) or (0,4).

What am I doing wrong here ?

Thank you in advance !

 

[Fermat1]

from which BC is "seen" at an angle of 90 degrees.

Can you make that precise?

 

[MarkFL]

Hello

I have two points: B(1,3) and C(2,6). I need to find a point A on the y-axis, from which BC is "seen" at an angle of 90 degrees.

I tried using Pythagoras theorem and got:

\[y^{2}-6y+10=y^{2}-12y+40+10\]

but it doesn't give the correct answer, which is (0,5) or (0,4).

What am I doing wrong here ?

Thank you in advance !

Here's what I would do:

Let the requested point be $P(0,y)$. Now we require segment $\overline{BP}$ to be normal to segment $\overline{CP}$, and so we require (given the product of the slopes of normal lines is -1):

$$\frac{3-y}{1-0}=\frac{0-2}{6-y}$$

$$(y-3)(y-6)=-2$$

$$y^2-9y+20=0$$

$$(y-4)(y-5)=0$$

Thus:

$$y\in\{4,5\}$$

 

[Yankel]

when I draw it, it looks like BP is normal to BC and not CP. This is why my answer is wrong. How could you tell which line is normal to which ?

 

[MarkFL]

when I draw it, it looks like BP is normal to BC and not CP. This is why my answer is wrong. How could you tell which line is normal to which ?

The line segments from our point $P$ on the $y$-axis to the two given points must be normal to each other. :D

 

[Yankel]

While I was drawing, I didn't keep the same scale on the x-axis and y-axis, and this is why in my drawing the angle did not look like 90 degrees. I understand the error now.

Thank you !