Missing coordinate problem in triangle

[dhie_20]

Hi,

I'm new in this forum and consider a non-mathematical person.
I have a problem with missing coordinate problem as shown in here



Let A(3,1) and B(12,4) is a segment with dist(A,B) = 9.49
CAB and DAB are right-angled triangles. dist(A,C) = dist (A,D) = 1, and dist(B,C) and dist(B,D) = 9.54
I need to solve the mystery of finding the coordinates for C and D

Most of the resources I got to solve this problem have C, A and D on X-axis. Due to my lack of luck, I couldn't find the solution to my problem.
The location of C and D can be easily got using Geogebra, but honestly, I don't know how they do the magic.?

To obtain the points in Geogebra, here is what I do:
1. Create a perpendicular line L from segment AB at point A
2. Create a circle C with r=1 at A
3. Obtain intersection L with C

Are there any other methods using the value from triangles only?

I really appreciate your help

Cheers

 

[HallsofIvy]

The line through (3, 1) and (12, 4) has slope (4- 1)/(12- 3)= 3/9= 1/3. The equation of the line through those points is y= (1/3)(x- 3)+ 1. The line perpendicular to that has slope -3. The line perpendicular to that line through (3, 1) has equation y= -3(x- 3)+ 1= -3x+ 10.

There are two points on y= -3x+ 10 at distance 1 from (3, 1). The distance from (3, 1) to point (x, y) is \(\displaystyle \sqrt{(x- 3)^2+ (y- 1)^2}= 1\) from which we can get \(\displaystyle (x- 3)^2+ (y- 1)^2= 1\).

Since y= -3x+ 10, that is \(\displaystyle (x- 3)^2+ (-3x+ 10- 1)^2= (x- 3)^2+ (-3x+ 9)^2= (x- 3)^2+ (-3(x- 3))^2= (x- 3)^2+ 9(x- 3)^2= 10(x- 3)^2= 1\).

And that is easy to solve. Dividing both sides by 10, \(\displaystyle (x- 3)^2= \frac{1}{10}\). Taking the square root of both sides, \(\displaystyle x- 3= \pm\sqrt{\frac{1}{10}}= \pm\frac{1}{\sqrt{10}}= \pm\frac{\sqrt{10}}{10}\).

So \(\displaystyle x= 3\pm \frac{\sqrt{10}}{10}\).

If \(\displaystyle x= 3+ \frac{\sqrt{10}}{10}\) then \(\displaystyle y= -3x+ 10= -9- \frac{3\sqrt{10}}{10}+ 10= 1- \frac{3\sqrt{10}}{10}\).


If \(\displaystyle x= 3- \frac{\sqrt{10}}{10}\) then \(\displaystyle y= -3x+ 10= -9+\frac{3\sqrt{10}}{10}+ 10= 1+ 3\frac{\sqrt{10}}{10}\).

Point C is \(\displaystyle \left(3- \frac{\sqrt{10}}{10}, 1+ 3\frac{\sqrt{10}}{10}\right)\).

Point D is \(\displaystyle \left(3+ \frac{\sqrt{10}}{10}. 1- \frac{3\sqrt{10}}{10}\right)\).

 

[dhie_20]

Thank you for your awesome answer.
I tried to understand your solution little by little.

Could you please explain this part:
" The line perpendicular to that has slope -3. " --> how do you know this slope?

I understand how to obtain the line between point A-B: y=(1/3)x. m=(1/3)
Does it mean that a perpendicular line to AB has a slope of -(1/m)?

 

[Steven G]

Suppose you have two perpendicular lines. If the slope of one line is a/b, then the slope of the other line is -b/a.

 

[dhie_20]

Kudos for @HallsofIvy and @Jomo
The problem has been solved

However, instead of using polynomial formula (apologies if I was wrong), I use a proportion of length to determine the coordinates.
Due to my limited skills, it's quite hard for me to implement polynomial rules in coding


Since the line of AB is y=(1/3) x and perpendicular at A is y=-3x+10, I tried to make a projection of A 1 unit to the left and 1 unit to the right.
From this step, I can easily get R1 and R2 using the perpendicular line of y= -3x+10

After I obtain the location of R1 and R2, I will get the distance of AR1 = AR2 = 3.16
The proportion of AC to AR1 = 1 / 3.16.
Therefore, I can easily get the location of
Cx = 3 - (1/3.16) * (3-2) = 2.68
Cy = 1 + (1/3.16) * (4-1) = 1.95

The same rule applied to D

Thanks again for your help