what will be the circumcentre of a triangle whose sides are 3x-y+3=0,3x+4y+3=0andx+3y+11=0?

Answer 1

Do the steps:

1) Find the vertices of the triangle, which are the intersection points of the lines or equations.
2) Find the perpendicular bisectors equations of two sides of
the triangle.
3) Find the intersection point of the preceding two lines.

Executing the steps above yields the results below:

1)
a)By solving each equation for y in terms of x, we get:
y= 3x+3; y= -3/4 x -3/4; y= -x/3 -11/3.
b) By equating the right sides two at a time and solving, we get
the vertices of the triangle to be (-1,0),(-2,3),(7,-6).

2) The perpendicular bisector of a segment is perpendicular to it
which means its slope is -1/slope of segment line. It bisects
the segment in half also or it passes its middle.
a) Middles of two segments is :
(-1,0) & (-2,-3) is ((-1-2)/2,(0-3)/2) = (-3/2,-3/2) and the slope of the segment is 3 so -1/3 is the slope required.
Result: the prependicular bisector is y = -1/3(x-(-3/2))-3/2

b)Middles of two segments is :
(-1,0) & (7,-6) is ((-1+7)/2,(0-6)/2) = (3,-3) and the slope of the segment is -3/4 so 4/3 is the slope required.
Result: the prependicular bisector is y = 4/3(x-3)-3

3) The intersection of y=4/3(x-3)-3 & y = -1/3(x-(-3/2))-3/2
is the circumcentre and it is found to be (3,-3).

Answer 2

what will be the circumcentre of a triangle whose sides are
3x - y + 3 = 0, 3x + 4y + 3 = 0 and x + 3y + 11=0?

3x - y + 3 = 0
3x + 4y + 3 = 0

Point of intersection is (-1, 0)

3x - y + 3 = 0
x + 3y + 11=0

Point of intersection is (-2, -3)

3x + 4y + 3 = 0
x + 3y + 11=0

Point of intersection is (7, -6)

So the vertices of the triangle are (-1, 0), (-2, -3) and (7, -6)

Let circumcentre be C (h, k)

Then the circum circle is the locus of points such that they are equidistant from (h, k) and contain the vertices

ie (x - h)² + (y - k)² = r²

So (-1 - h)² + (0 - k)² = r² ... (1a)

ie h² + 2h + 1 + k² = r² ... (1)

(-2 - h)² + (-3 - k)² = r² ... (2a)

ie h² + 4h + 4 + k² + 6k + 9 = r² ... (2)

(7 - h)² + (-6 - k)² = r² ... (3a)

ie h² - 14h + 49 + k² + 12k + 36 = r² ... (3)

Equation (2) - equation (1)

2h + 3 + 6k + 9 = 0
ie h + 3k = -6 ... (4)

Equation (3) - equation (2)

-18h + 45 + 6k + 27 = 0

ie 3h - k = 12 ... (5)

Equation (4) + 3 x equation (5)

10h = 30

So h = 3, thus k = -3, r² = 25 so r = 5

Solution is h = 3 k = -3 and r = 5

So the circumcentre is (3, -3) and the radius of the circumcircle is 5

Answer 3

3x - y + 3 = 0, 3x + 4y + 3 = 0 and x + 3y + 11=0?

3x - y + 3 = 0
3x + 4y + 3 = 0

Point of intersection is (-1, 0)

3x - y + 3 = 0
x + 3y + 11=0

Point of intersection is (-2, -3)

3x + 4y + 3 = 0
x + 3y + 11=0

Point of intersection is (7, -6)

So the vertices of the triangle are (-1, 0), (-2, -3) and (7, -6)

Let circumcentre be C (h, k)

Then the circum circle is the locus of points such that they are equidistant from (h, k) and contain the vertices

ie (x - h)² + (y - k)² = r²

So (-1 - h)² + (0 - k)² = r² ... (1a)

ie h² + 2h + 1 + k² = r² ... (1)

(-2 - h)² + (-3 - k)² = r² ... (2a)

ie h² + 4h + 4 + k² + 6k + 9 = r² ... (2)

(7 - h)² + (-6 - k)² = r² ... (3a)

ie h² - 14h + 49 + k² + 12k + 36 = r² ... (3)

Equation (2) - equation (1)

2h + 3 + 6k + 9 = 0
ie h + 3k = -6 ... (4)

Equation (3) - equation (2)

-18h + 45 + 6k + 27 = 0

ie 3h - k = 12 ... (5)

Equation (4) + 3 x equation (5)

10h = 30

So h = 3, thus k = -3, r² = 25 so r = 5

Solution is h = 3 k = -3 and r = 5

So the circumcentre is (3, -3) and the radius of the circumcircle is 5

Answer 4

A triangle defined by the lines
3x - y + 3 = 0,
3x + 4y + 3 = 0
x + 3y + 11 = 0

3x - y = - 3
3x + 4y = - 3
3x + 9y = - 33
Solving pairwise for vertices.
3x + 4y = - 3
3x - y = - 3
1) y = 0, x = -1 (3,-3)◄
3x + 9y = - 33
-3x - 4y = + 3
2) y = -6, x = 7 (5/2,-9/2)◄
3x + 9y = - 33
-3x + y = + 3
3) y = -3, x = -2 (-3/2, -3/2)◄
Midpoints are:
1) (3,-3)◄3x + 4y = - 3, y = (-3/4)x - 3/4
2) (5/2,-9/2)◄ x + 3y = - 11, y = (-1/3)x - 11/3
3) (-3/2, -3/2)◄ 3x - y = - 3, y = -3x + 3
Two orthogonal lines are:
(y + 3) = (4/3)(x - 3)
(y + 9/2) = 3(x - 5/2)
3y + 9 = 4x - 12
2y + 9 = 6x - 15
-6y - 18 = -8x + 24
6y + 27 = 18x - 45
9 = 10x - 21
10x = 30
x = 3
2y + 9 = 18 - 15
2y = -6
y = -3
Circumcenter = (3, -3)
(-1, 0)
r = √4^2 + 3^2)
r = √25 = 5
(7, -6)
r = √4^2 + 3^2 = 5

Answer 5

There are several ways to solve this. One of the ways is to determine the points of intersection of these three lines.

y + 3 = 0
3x + 4y = 0
x + 3y + 11 = 0

First, let's put them all into slope-intercept form.

L1: y = -3
L2: y = (-3/4)x
L3: y = (-1/3)x - 11/3

Solve for the intersection between L1 and L2:

-3 = (-3/4)x
-12 = -3x
x = 4, therefore y = -3.

First point: (4, -3)

Intersection between L2 and L3:

y = (-3/4)x
y = (-1/3)x - 11/3
Therefore,

(-3/4)x = (-1/3)x - 11/3, Multiplying both sides by 12, to get
(-3)(3)x = (-4)x - 11(4)
-9x = -4x - 44
-5x = -44
x = 44/5

Since x = 44/5, y = (-3/4)(44/5) = -33/5

Second point: (44/5, -33/5)

Intersection between L1 and L3:

L1: y = -3
L3: y = (-1/3)x - 11/3

Equate both of them.
-3 = (-1/3)x - 11/3. Multiply both sides by 3, to get
-3 = (-1)x - 11
-3 = -x - 11
x = -11 + 3
x = -8

If x = -8, y = -3

Third point: (-8, -3)

Three vertices of the triangle are at: (4, -3), (44/5, -33/5), (-8, -3).

All we have to do is interpolate these points in the form of a circle, and obtain the coordinates of the center of the circle.

The formula for a circle goes as follows:

(x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center, and r is the radius.

But since we have these coordinates (4, -3), (44/5, -33/5), (-8, -3), we just just plug these in for x and y respectively.

Let's start with (4,-3).

(x - h)^2 + (y - k)^2 = r^2
(4 - h)^2 + (-3 - k)^2 = r^2
[h^2 - 8h + 16] + [k^2 + 6k + 9] = r^2

And I find that the problem has become way too long. I must cut this short, I'm sorry.

Answer 6

3x - y = 5 and x + 3y = 5 i began with the small numbers so i tried x = 2 and y =a million. That worked so I went to the subsequent one. 2x - 3y = 3 and x + 4y = 7 I appeared on the 2d of those sums and realised that y should be a million because if it replaced into any larger, than 7 replaced into out of attain. (until eventually negatives are in this, yet i have not performed algebra with negatives beforehand.) So y = a million back and x + (4x1) = 7 would make x 3. (2x3) - (3x1) = 3 also so the answer replaced into shown.

Answer 7

The 3 formulas represent 3 lines that are the sides of the triangle. So graph the triangle. After graphing the triangle, you will be able to visualize the geometric solution to computing the circumcentre of the triange. By the way, a circumcentre is figured out as explained in the website in the Source.