Why am I getting different incentres for the same triangle?

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Maths Pro

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The question reads : O(0,0), A(6,0) and B(0,8) are the vertices of a triangle. Find the coordinates of the incenter of triangle OAB.
I tried to find equation of AB using two-point form. I get equation as 4x + 3y- 24 = 0. Now since mod of length of perpendicular dropped from incentre(assumed as P(a, b)) to this line is given by mod [(ax+by+c)/root(a^2 +b^2)] = p(perpendicular length), I get mod[(4a+3b-24)/root(4^2 +3^2)] = 2*****
implies mod [(4a+3b-24)/root(16+9))] = 2
implies mod[(4a+3b-24)/root 25] = 2
implies mod[(4a+3b-24)/5] = 2
implies mod [(4a+3b-24] = 10(assuming positive value of R.H.S)
or 4a+3b = 34....(1). Now equation of angle bisector of angle BOC is y = x, (since this line has slope tan 45=1). Line y = x passes through P(a, b)
Hence b=a or a-b =0....(2)
Solving (1) and (2) simultaneously, I get a=34/7 and hence b=34/7*** Or incentre as (34/7, 34/7). Answer is (2,2).
***** I got this length of perpendicular 2 by this method :
Distance from (foot of perpendicular of side OA from P(a, b)) to point A(6,0)= 6-m and distance of point O(0,0) to (foot of perpendicular of side OA from point P(a, b)) = m. Similarly, distance from (foot of perpendicular of side OB from P(a, b)) to point B(0,8)= 8-m and distance of point O(0,0) to (foot of perpendicular of side OB from point P(a, b)) = m.
Length of AB(hypotenuse of OAB)= (8-m)+(6-m) =14-2m=10 => m = 2
So here itself I get radius of incentre as 2 and hence the coordinates of incentre as (2,2)
But then why this anomaly here***
What's happening here?
I am sorry for this long question and thanks for reading it patiently.
 
If by mod[4a+3b-24] you mean [imath]\lvert 4a+3b-24 \rvert[/imath] normally called absolute value, then you need to consider that the result could be [imath]\pm 10[/imath]. Using [imath]-10[/imath] will give you the correct answer. To understand why, look up the formula for distance from point to line. It is a directed formula meaning that the point you want [imath](a,b)[/imath] will either be "above" or "below" the line depending on sign.
 
mrtwhs said:
If by mod[4a+3b-24] you mean ∣4a+3b−24∣\lvert 4a+3b-24 \rvert∣4a+3b−24∣ normally called absolute value, then you need to consider that the result could be ±10\pm 10±10. Using −10-10−10 will give you the correct answer. To understand why, look up the formula for distance from point to line. It is a directed formula meaning that the point you want (a,b)(a,b)(a,b) will either be "above" or "below" the line depending on sign.
Click to expand...
But how would I know which sign to apply. There is incentre formula but I don't have it's proof.
 
It would have helped in communicating your work if you had made a picture (and labeled the various points you referred to); that will also help you understand it.

You found that |4a+3b-24| = 10; this means that either 4a+3b-24 = 10 (so that 4a+3b = 34) or 4a+3b-24 = -10 (so that 4a+3b = 14). These are two lines, each 2 units away from AB:

1770571817046.png​

You used the green line, and found a point on the wrong side of AB (D instead of C):

1770571929314.png​

That's a deficiency in your method; but it's not hard to see how to decide which to use. (Use the line closer to O, which means the smaller number on the RHS.)
Maths Pro said:
There is incentre formula but I don't have it's proof.
Click to expand...
What is the formula you know?
 
This isn't exactly a universal formula but with [imath]O(0,0)[/imath], [imath]A(6,0)[/imath], [imath]B(0,8)[/imath] and [imath]I(r,r)[/imath], you can use the formula [imath]A=rs[/imath] where [imath]A=[/imath] area, [imath]r=[/imath] inradius, and [imath]s=[/imath] semi-perimeter. This leads to the equation [imath]24=12r[/imath] and so [imath]r=2[/imath].

It might be worthwhile looking up barycentric coordinates. Ordering the vertices [imath]O,A,B[/imath] the barycentric coordinates are [imath](10:8:6)[/imath]. Normalizing them yields [imath](10/24,8/24,6/24)=(5/12,1/3,1/4)[/imath]. The coordinates can be found by converting these back to cartesian coordinates [imath](5/12) \cdot 0+(1/3) \cdot 6 +(1/4) \cdot 0=2[/imath] and [imath](5/12) \cdot 0+(1/3) \cdot 0 +(1/4) \cdot 8=2[/imath].
 
We are given triangle vertices:
To find the incentre, we use the formula:
where a, b c are the side lengths opposite to vertices A respectively.
 
Oh sorry, here it is :
Incentre (ax1+bx2+cx3)/(a+b+c) , (ay1+by2+cy3)/(a+b+c))
Where a, b,c are sides opposite to vertices A,B and O respectively. O(0,0), A(0,6) and B(0,8). Is it using section formula?
 
Maths Pro said:
Oh sorry, here it is :
Incentre (ax1+bx2+cx3)/(a+b+c) , (ay1+by2+cy3)/(a+b+c))
Where a, b,c are sides opposite to vertices A,B and O respectively. O(0,0), A(0,6) and B(0,8). Is it using section formula?
Click to expand...
A close examination will show you that this is essentially the same as my barycentric coordinate solution.
 
Maths Pro said:
Do you have proof for that? Please share
Click to expand...
Given [imath]\triangle ABC[/imath] with barycentric coordinates [imath]A(1,0,0)[/imath], [imath]B(0,1,0)[/imath], [imath]C(0,0,1)[/imath] then knowing how an angle bisector divides the side of the triangle that it hits allows you to quickly get [imath]I(a,b,c)[/imath] as the incenter. To normalize, divide all three coordinates by [imath]a+b+c[/imath] so that the sum of the coefficients is [imath]1[/imath]. I'll leave it to you to look up how to convert barycentric coordinates to cartesian coordinates. It is essentially what you did in post #9.
 
Maths Pro said:
The question reads : O(0,0), A(6,0) and B(0,8) are the vertices of a triangle. Find the coordinates of the incenter of triangle OAB.
I tried to find equation of AB using two-point form. I get equation as 4x + 3y- 24 = 0. Now since mod of length of perpendicular dropped from incentre(assumed as P(a, b)) to this line is given by mod [(ax+by+c)/root(a^2 +b^2)] = p(perpendicular length), I get mod[(4a+3b-24)/root(4^2 +3^2)] = 2*****
implies mod [(4a+3b-24)/root(16+9))] = 2
implies mod[(4a+3b-24)/root 25] = 2
implies mod[(4a+3b-24)/5] = 2
implies mod [(4a+3b-24] = 10(assuming positive value of R.H.S)
or 4a+3b = 34....(1). Now equation of angle bisector of angle BOC is y = x, (since this line has slope tan 45=1). Line y = x passes through P(a, b)
Hence b=a or a-b =0....(2)
Solving (1) and (2) simultaneously, I get a=34/7 and hence b=34/7*** Or incentre as (34/7, 34/7). Answer is (2,2).
***** I got this length of perpendicular 2 by this method :
Distance from (foot of perpendicular of side OA from P(a, b)) to point A(6,0)= 6-m and distance of point O(0,0) to (foot of perpendicular of side OA from point P(a, b)) = m. Similarly, distance from (foot of perpendicular of side OB from P(a, b)) to point B(0,8)= 8-m and distance of point O(0,0) to (foot of perpendicular of side OB from point P(a, b)) = m.
Length of AB(hypotenuse of OAB)= (8-m)+(6-m) =14-2m=10 => m = 2
So here itself I get radius of incentre as 2 and hence the coordinates of incentre as (2,2)
But then why this anomaly here***
What's happening here?
I am sorry for this long question and thanks for reading it patiently.
Click to expand...
Can you "derive" the co-ordinates of the incenter of a triangle, whose length of the sides are a, b & c?
 
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