1. Circles and angle bisectors

I got a problem with solving this problem on circles....
The sides of a triangle lie on the lines 3x+y-5=0, x - 3y-1 = 0 and x+3y+7=0 find the equation of the circle inscribed in the triangle..

Helppp ... How am i going to use angle bisector in this problem?

(x-\frac{5}{6})^2 + (y+\frac{4}{3})^2 = \frac{529}{360}
Thanks and Merry Christmas All!

Btw hows latex doin is broken again?

I got a problem with solving this problem on circles....
The sides of a triangle lie on the lines 3x+y-5=0, x - 3y-1 = 0 and x+3y+7=0 find the equation of the circle inscribed in the triangle..

Helppp ... How am i going to use angle bisector in this problem?

(x-\frac{5}{6})^2 + (y+\frac{4}{3})^2 = \frac{529}{360}
Thanks and Merry Christmas All!

Btw hows latex doin is broken again?
I have an idea how to do it.
Are you familiar with the formula for two lines (not perpendicular)?
\tan \theta=\frac{m_1-m_2}{1+m_1m_2}
Where,
m_1,m_2
Are the slopes of the two lines.
~~~~~~
Here is an example of what I said.
~~~~~~
The first line you gave was,
3x+y-5=0
The second line you gave was,
x-3y-1=0
Thus,
y=-3x+5
y=(1/3)x-(1/3)
Thus, the angle between them is, (note the negative)
tan \theta = \frac{ 1/3+3}{ 1+(1/3)(-3) }
Notice the numerator is zero () because they are perpendicular. Thus, you now the angle between them is 90 degrees.

Thus, the angle of the bisector needs to do is 45 degree between these two lines.
Let the slope be "m" thus, (between the lower line)
\tan 45 = \frac{ m - 1/3 }{1+m/3}
Thus,
1= \frac{ m - 1/3 }{1 +m/3}
Thus,
1+m/3=m-1/3
Multiply through by 3,
3+m=3m-1
4=2m
2=m

Thus, the slope of the angle bisector is 2.

Now we find the point of intersection (because we already have the slope, if we have a point on a line then we can find its equation).

We do that by solving the equation,
y=-3x+5
y=(1/3)x-(1/3)
Thus,
-3x+5=(1/3)x-(1/3)
Multiply by 3,
-9x+15=x-1
16=10x
Thus,
x=1.6
y=-3(1.6)+5=.2

We will use the point intercept formula,
The equation of a line (non-vertical) is,
y-y_0=m(x-x_0)
Where, "m" is the slope and, (x_0,y_0) is a point on line.

Thus,
y-.2=2(x-1.6)
y-.2=2x-3.2
y=2x-3

3. But wait we are not other.
There really is a lot to do in this problem.

Pick another pair of lines and find the equation of angle bisector like I did before.

Now, the trick is not to find the equation of the last angle bisector. Because it intersects in the exact same point as the other two.

Thus, after you have the two angle bisectors find where they intersect (by solving a linear system). Thus, you know the center of the circle!

But how do we now the radius? The radius of the circle?
One way it to use the distance formula for a point and line.
But I have a much cooler way.

Use the following useful theorem in geometry,

Theorem: The radius of the inscribed circle is the ratio of the area of the triangle to the semi-perimeter.

The semi-perimeter, is found by find the perimeter (finding the sum of all the lengths of find) and dividind it by 2.

The area is found by a famous theorem called, "Heron's formula":
A=\sqrt{ s(s-a)(s-b)(s-c) }
Where,
s---> semiperimeter
a,b,c---> sides of triangle.

Thus, you next step now, after finding the center, is to find the radius. Which is done first by finding the length of sides of the triangle.

...
The sides of a triangle lie on the lines 3x+y-5=0, x - 3y-1 = 0 and x+3y+7=0 find the equation of the circle inscribed in the triangle..

Helppp ... How am i going to use angle bisector in this problem?

...
Hello,
Merry Christmas!

If you have to use angle bisectors then TPH has already described a possible way to solve the problem.

I'll give you a hint how to do this problem by using parallels of the sides that means of the given lines: The centre of the circle is the intercept of three lines (see attachment) which have always the same distance (r) to the lines. Thus you'll get a system of three equations to solve for the coordinates of the circle and the radius.

EB

I have a solution which avoids angle bisectors,
. . but it's still quite long and involved.

The sides of a triangle lie on the lines
. . A: 3x + y - 5 = 0
. . B: x - 3y - 1 = 0
. . C: x + 3y + 7 = 0

Find the equation of the circle inscribed in the triangle..

Answer at back of book: .(x - 5/6)² + (y + 4/3)² = 529/60

I sketched the lines.
The incenter is in quadrant 4, where x is positive, y is negative.

Distance from a point to a line

Given point P(x1,y1) and line L: Ax + By + C = 0, the distance from P to L is:

. . . . . . . |A(x1) + B(y1) + C|
. . d . = . ------------------------
. . . . . . . . . . √A² + B²

We want the center P(h,k) which is equidistant from A, B, and C.

. . . . . |3h + k - 5|
PA .= .--------------
. . . . . . . .√10

. . . . . |h - 3k - 1|
PB .= .--------------
. . . . . . . .√10

. . . . . |h + 3k + 7|
PC .= .--------------
. . . . . . . .√10

The numerator of PA is negative.
. . Hence, we have: . -3h - k + 5 . = . h - 3k - 1 . = . h + 3k + 7

From the last two: . h - 3k - 1 .= .h + 3k + 7 . . -6k = 8 . . k = -4/3
. . Then: .h = 5/6

Hence, the center is: . P(5/6, -4/3)

The radius is the distance from P to line B:

. . . . . .|(5/6) - 3(-4/3) - 1| . . . .23
. . r .= .----------------------- .= .-------
. . . . . . . . . . .√10 . . . . . . . . . 6√10

Therefore: .(x - h)² + (y - k)² .= .

. . gives us: . (x - 5/6)² + (y + 4/3)² .= .529/360

Whew! . . . I need a nap.

6. Good work im almost near solving this thanks to you all

Hello soroban thanks very much
ive got only one problem
how did you get 5/6? because i cant really get it solving the
3 lines = 0 w/ k = -4/3

And once again thanks vry much for all the help

Good work im almost near solving this thanks to you all

Hello soroban thanks very much
ive got only one problem
how did you get 5/6? because i cant really get it solving the
3 lines = 0 w/ k = -4/3
And once again thanks vry much for all the help
Hello,

take the first equation from

-3h - k + 5 = h - 3k - 1 = h + 3k + 7 You'll get:

-3h - k + 5 = h - 3k - 1 Now plug in the value k = (-4/3)

-3h - (-4/3) + 5 = h - 3*(-4/3) - 1 and solve for h:

-4h = -19/3 + 4 - 1

h = 10/12 = 5/6

EB