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5. Find Centroid, Circumcenter, Area of a triangle example
( Enter your problem )
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- Find the centroid of a triangle whose vertices are A(4,-6),B(3,-2),C(5,2)
- Find the circumcentre of a triangle whose vertices are A(-2,-3),B(-1,0),C(7,-6)
- Using determinants, find the area of the triangle with vertices are A(-3,5),B(3,-6),C(7, 2)
- Using determinants show that the following points are collinear A(2,3),B(-1,-2),C(5,8)
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Other related methods
- Distance, Slope of two points
- Points are Collinear or Triangle or Quadrilateral form
- Find Ratio of line joining AB and is divided by P
- Find Midpoint or Trisection points or equidistant points on X-Y axis
- Find Centroid, Circumcenter, Area of a triangle
- Find the equation of a line using slope, point, X-intercept, Y-intercept
- Find Slope, X-intercept, Y-intercept of a line
- Find the equation of a line passing through point of intersection of two lines and slope or a point
- Find the equation of a line passing through a point and parallel or perpendicular to Line-2 or point-2 and point-3
- Find the equation of a line passing through point of intersection of Line-1, Line-2 and parallel or perpendicular to Line-3
- For two lines, find Angle, intersection point and determine if parallel or perpendicular lines
- Reflection of points about x-axis, y-axis, origin
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1. Find the centroid of a triangle whose vertices are A(4,-6),B(3,-2),C(5,2)
(Previous example) | 3. Using determinants, find the area of the triangle with vertices are A(-3,5),B(3,-6),C(7, 2)
(Next example) |
2. Find the circumcentre of a triangle whose vertices are A(-2,-3),B(-1,0),C(7,-6)
1. Find the circumcentre of a triangle whose vertices are `A(-2,-3),B(-1,0),C(7,-6)`
Solution:
The center of a circle which passes through all the vertices of a triangle is called circumcentre of a triangle.
Vertices of triangle are `A(-2,-3),B(-1,0)` and `C(7,-6)`
Let `P(x,y)` be the circumcentre of a triangle
`:. PA=PB=PC`
`:. PA^2=PB^2=PC^2`
Now `PA^2=PB^2`
`(x+2)^2+(y+3)^2=(x+1)^2+(y-0)^2`
`(x^2+4x+4)+(y^2+6y+9)=(x^2+2x+1)+(y^2)`
`2x+6y=-12`
`x+3y=-6 ->(1)`
Now `PB^2=PC^2`
`(x+1)^2+(y-0)^2=(x-7)^2+(y+6)^2`
`(x^2+2x+1)+(y^2)=(x^2-14x+49)+(y^2+12y+36)`
`16x-12y=84`
`4x-3y=21 ->(2)`
Intersection point of equation `(1)` and `(2)`
The point of intersection of the lines can be obtainted by solving the given equations
`x+3y=-6`
and `4x-3y=21`
`x+3y=-6 ->(1)`
`4x-3y=21 ->(2)`
Adding `=>5x=15`
`=>x=15/5`
`=>x=3`
Putting `x=3` in equation `(1)`, we have
`3+3y=-6`
`=>3y=-6-3`
`=>3y=-9`
`=>y=-3`
`:.x=3" and "y=-3`
`:. (3,-3)` is the intersection point of the given two lines.
`:.` Circumcentre of a triangle is `P(3,-3)`
Radius of the Circumcircle
`AP=sqrt((3+2)^2+(-3+3)^2)=sqrt((5)^2+(0)^2)=sqrt(25+0)=sqrt(25)=5`
2. Find the circumcentre of a triangle whose vertices are `A(-1,0),B(-1,2),C(3,2)`
Solution:
The center of a circle which passes through all the vertices of a triangle is called circumcentre of a triangle.
Vertices of triangle are `A(-1,0),B(-1,2)` and `C(3,2)`
Let `P(x,y)` be the circumcentre of a triangle
`:. PA=PB=PC`
`:. PA^2=PB^2=PC^2`
Now `PA^2=PB^2`
`(x+1)^2+(y-0)^2=(x+1)^2+(y-2)^2`
`(x^2+2x+1)+(y^2)=(x^2+2x+1)+(y^2-4y+4)`
`+4y=4`
`+y=1 ->(1)`
Now `PB^2=PC^2`
`(x+1)^2+(y-2)^2=(x-3)^2+(y-2)^2`
`(x^2+2x+1)+(y^2-4y+4)=(x^2-6x+9)+(y^2-4y+4)`
`8x=8`
`x=1 ->(2)`
Intersection point of equation `(1)` and `(2)`
The point of intersection of the lines can be obtainted by solving the given equations
`+y=1`
`:.y=1`
and `x=1`
`y=1 ->(1)`
`x=1 ->(2)`
``
`:. (1,1)` is the intersection point of the given two lines.
`:.` Circumcentre of a triangle is `P(1,1)`
Radius of the Circumcircle
`AP=sqrt((1+1)^2+(1-0)^2)=sqrt((2)^2+(1)^2)=sqrt(4+1)=sqrt(5)=sqrt(5)`
3. Find the circumcentre of a triangle whose vertices are `A(-2,3),B(2,-1),C(4,0)`
Solution:
The center of a circle which passes through all the vertices of a triangle is called circumcentre of a triangle.
Vertices of triangle are `A(-2,3),B(2,-1)` and `C(4,0)`
Let `P(x,y)` be the circumcentre of a triangle
`:. PA=PB=PC`
`:. PA^2=PB^2=PC^2`
Now `PA^2=PB^2`
`(x+2)^2+(y-3)^2=(x-2)^2+(y+1)^2`
`(x^2+4x+4)+(y^2-6y+9)=(x^2-4x+4)+(y^2+2y+1)`
`8x-8y=-8`
`x-y=-1 ->(1)`
Now `PB^2=PC^2`
`(x-2)^2+(y+1)^2=(x-4)^2+(y-0)^2`
`(x^2-4x+4)+(y^2+2y+1)=(x^2-8x+16)+(y^2)`
`4x+2y=11 ->(2)`
Intersection point of equation `(1)` and `(2)`
The point of intersection of the lines can be obtainted by solving the given equations
`x-y=-1`
and `4x+2y=11`
`x-y=-1 ->(1)`
`4x+2y=11 ->(2)`
equation`(1) xx 2 =>2x-2y=-2`
equation`(2) xx 1 =>4x+2y=11`
Adding `=>6x=9`
`=>x=9/6`
`=>x=3/2`
Putting `x=3/2 ` in equation `(1)`, we have
`(3/2)-y=-1`
`=>-y=-1-(3/2)`
`=>-y=(-2-3)/2`
`=>-y=-5/2`
`=>y=5/2`
`:.x=3/2" and "y=5/2`
`:. (3/2,5/2)` is the intersection point of the given two lines.
`:.` Circumcentre of a triangle is `P(3/2,5/2)`
Radius of the Circumcircle
`AP=sqrt((3/2+2)^2+(5/2-3)^2)=sqrt((7/2)^2+(-1/2)^2)=sqrt(49/4+1/4)=sqrt(25/2)=5/(sqrt(2))`
4. Find the circumcentre of a triangle whose vertices are `A(1,3),B(-3,5),C(5,-1)`
Solution:
The center of a circle which passes through all the vertices of a triangle is called circumcentre of a triangle.
Vertices of triangle are `A(1,3),B(-3,5)` and `C(5,-1)`
Let `P(x,y)` be the circumcentre of a triangle
`:. PA=PB=PC`
`:. PA^2=PB^2=PC^2`
Now `PA^2=PB^2`
`(x-1)^2+(y-3)^2=(x+3)^2+(y-5)^2`
`(x^2-2x+1)+(y^2-6y+9)=(x^2+6x+9)+(y^2-10y+25)`
`-8x+4y=24`
`2x-y=-6 ->(1)`
Now `PB^2=PC^2`
`(x+3)^2+(y-5)^2=(x-5)^2+(y+1)^2`
`(x^2+6x+9)+(y^2-10y+25)=(x^2-10x+25)+(y^2+2y+1)`
`16x-12y=-8`
`4x-3y=-2 ->(2)`
Intersection point of equation `(1)` and `(2)`
The point of intersection of the lines can be obtainted by solving the given equations
`2x-y=-6`
and `4x-3y=-2`
`2x-y=-6 ->(1)`
`4x-3y=-2 ->(2)`
equation`(1) xx 3 =>6x-3y=-18`
equation`(2) xx 1 =>4x-3y=-2`
Substracting `=>2x=-16`
`=>x=-16/2`
`=>x=-8`
Putting `x=-8` in equation `(1)`, we have
`2(-8)-y=-6`
`=>-y=-6+16`
`=>-y=10`
`=>y=-10`
`:.x=-8" and "y=-10`
`:. (-8,-10)` is the intersection point of the given two lines.
`:.` Circumcentre of a triangle is `P(-8,-10)`
Radius of the Circumcircle
`AP=sqrt((-8-1)^2+(-10-3)^2)=sqrt((-9)^2+(-13)^2)=sqrt(81+169)=sqrt(250)=5sqrt(10)`
5. Find the circumcentre of a triangle whose vertices are `A(1,3),B(0,-2),C(-3,1)`
Solution:
The center of a circle which passes through all the vertices of a triangle is called circumcentre of a triangle.
Vertices of triangle are `A(1,3),B(0,-2)` and `C(-3,1)`
Let `P(x,y)` be the circumcentre of a triangle
`:. PA=PB=PC`
`:. PA^2=PB^2=PC^2`
Now `PA^2=PB^2`
`(x-1)^2+(y-3)^2=(x-0)^2+(y+2)^2`
`(x^2-2x+1)+(y^2-6y+9)=(x^2)+(y^2+4y+4)`
`-2x-10y=-6`
`x+5y=3 ->(1)`
Now `PB^2=PC^2`
`(x-0)^2+(y+2)^2=(x+3)^2+(y-1)^2`
`(x^2)+(y^2+4y+4)=(x^2+6x+9)+(y^2-2y+1)`
`-6x+6y=6`
`x-y=-1 ->(2)`
Intersection point of equation `(1)` and `(2)`
The point of intersection of the lines can be obtainted by solving the given equations
`x+5y=3`
and `x-y=-1`
`x+5y=3 ->(1)`
`x-y=-1 ->(2)`
Substracting `=>6y=4`
`=>y=4/6`
`=>y=2/3`
Putting `y=2/3 ` in equation `(2)`, we have
`x-(2/3)=-1`
`=>x=-1+(2/3)`
`=>x=(-3+2)/3`
`=>x=-1/3`
`:.x=-1/3" and "y=2/3`
`:. (-1/3,2/3)` is the intersection point of the given two lines.
`:.` Circumcentre of a triangle is `P(-1/3,2/3)`
Radius of the Circumcircle
`AP=sqrt((-1/3-1)^2+(2/3-3)^2)=sqrt((-4/3)^2+(-7/3)^2)=sqrt(16/9+49/9)=sqrt(65/9)=(sqrt(65))/3`
6. Find the circumcentre of a triangle whose vertices are `A(1,2),B(3,-4),C(5,-6)`
Solution:
The center of a circle which passes through all the vertices of a triangle is called circumcentre of a triangle.
Vertices of triangle are `A(1,2),B(3,-4)` and `C(5,-6)`
Let `P(x,y)` be the circumcentre of a triangle
`:. PA=PB=PC`
`:. PA^2=PB^2=PC^2`
Now `PA^2=PB^2`
`(x-1)^2+(y-2)^2=(x-3)^2+(y+4)^2`
`(x^2-2x+1)+(y^2-4y+4)=(x^2-6x+9)+(y^2+8y+16)`
`4x-12y=20`
`x-3y=5 ->(1)`
Now `PB^2=PC^2`
`(x-3)^2+(y+4)^2=(x-5)^2+(y+6)^2`
`(x^2-6x+9)+(y^2+8y+16)=(x^2-10x+25)+(y^2+12y+36)`
`4x-4y=36`
`x-y=9 ->(2)`
Intersection point of equation `(1)` and `(2)`
The point of intersection of the lines can be obtainted by solving the given equations
`x-3y=5`
and `x-y=9`
`x-3y=5 ->(1)`
`x-y=9 ->(2)`
Substracting `=>-2y=-4`
`=>2y=4`
`=>y=4/2`
`=>y=2`
Putting `y=2` in equation `(2)`, we have
`x-(2)=9`
`=>x=9+2`
`=>x=11`
`:.x=11" and "y=2`
`:. (11,2)` is the intersection point of the given two lines.
`:.` Circumcentre of a triangle is `P(11,2)`
Radius of the Circumcircle
`AP=sqrt((11-1)^2+(2-2)^2)=sqrt((10)^2+(0)^2)=sqrt(100+0)=sqrt(100)=10`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
1. Find the centroid of a triangle whose vertices are A(4,-6),B(3,-2),C(5,2)
(Previous example) | 3. Using determinants, find the area of the triangle with vertices are A(-3,5),B(3,-6),C(7, 2)
(Next example) |
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