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Method and examples
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Method
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Circumcenter of a triangle
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1. Distance, Slope of two points
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1. Find the distance between the points `A(5,-8)` and `B(-7,-3)`
2. Find the slope of the line joining points `A(4,-8)` and `B(5,-2)`
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` - `A(5,-8),B(-7,-3)`
- `A(7,-4),B(-5,1)`
- `A(-6,-4),B(9,-12)`
- `A(1,-3),B(4,-6)`
- `A(-5,7),B(-1,3)`
- `A(-8,6),B(2,0)`
- `A(0,0),B(7,4)`
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Find the value of x or y
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3. If distance between the points (5,3) and (x,-1) is 5, then find the value of x.
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Distance =
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- `A(5,3),B(x,-1)`, distance `=5`
- `A(x,-1),B(3,2)`, distance `=5`
- `A(x,2),B(3,-6)`, distance `=10`
- `A(x,1),B(-1,5)`, distance `=5`
- `A(x,7),B(1,15)`, distance `=10`
- `A(1,x),B(-3,5)`, distance `=5`
- `A(x,0),B(4,8)`, distance `=10`
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4. If slope of the line joining points `A(x,0), B(-3,-2)` is `2/7`, find the value of `x`
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Slope =
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- `A(x,0),B(-3,-2)`, slope `=2/7`
- `A(2,x),B(-3,7)`, slope `=1`
- `A(x,5),B(-1,2)`, slope `=3/4`
- `A(2,5),B(x,3)`, slope `=2`
- `A(x,2),B(6,-8)`, slope `=-5/4`
- `A(-2,x),B(5,-7)`, slope `=-1`
- `A(2,3),B(x,6)`, slope `=3/5`
- `A(-3,4),B(5,x)`, slope `=-5/4`
- `A(0,x),B(5,-2)`, slope `=-9/5`
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2. Points are Collinear or Triangle or Quadrilateral form
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Show that the points are the vertices of
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Find `A(0,0), B(2,2), C(0,4), D(-2,2)` are vertices of a square or not
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- `A(1,5),B(2,3),C(-2,-11)` are collinear points
- `A(1,-3),B(2,-5),C(-4,7)` are collinear points
- `A(-1,-1),B(1,5),C(2,8)` are collinear points
- `A(0,-1),B(3,5),C(5,9)` are collinear points
- `A(2,8),B(1,5),C(0,2)` are collinear points
- `A(-1,-1),B(1,5),C(2,8)` are collinear points
- `A(0,-1),B(3,5),C(5,9)` are collinear points
- `A(2,8),B(1,5),C(0,2)` are collinear points
- `A(0,0),B(0,3),C(4,0)` are vertices of a right angle triangle
- `A(-2,-2),B(-1,2),C(3,1)` are vertices of a right angle triangle
- `A(-3,2),B(1,2),C(-3,5)` are vertices of a right angle triangle
- `A(2,5),B(8,5),C(5,10.196152)` are vertices of an equilateral triangle
- `A(2,2),B(-2,4),C(2,6)` are vertices of an isosceles triangle
- `A(0,0),B(2,0),C(-4,0),D(-2,0)` are collinear points
- `A(3,2),B(5,4),C(3,6),D(1,4)` are vertices of a square
- `A(0,0),B(2,2),C(0,4),D(-2,2)` are vertices of a square
- `A(1,-1),B(-2,2),C(4,8),D(7,5)` are vertices of a rectangle
- `A(0,-4),B(6,2),C(3,5),D(-3,-1)` are vertices of a rectangle
- `A(3,0),B(4,5),C(-1,4),D(-2,-1)` are vertices of a rhombus
- `A(2,3),B(7,4),C(8,7),D(3,6)` are vertices of a parallelogram
- `A(1,5),B(1,4),C(-1,3),D(-1,4)` are vertices of a parallelogram
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3. Find Ratio of line joining AB and is divided by P
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1. Find the ratio in which the point P(3/4, 5/12) divides the line segment joining the points A(1/2, 3/2) and B(2, -5)
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- `P(3/4,5/12),A(1/2,3/2),B(2,-5)`
- `P(-1,6),A(3,10),B(6,-8)`
- `P(-2,3),A(-3,5),B(4,-9)`
- `P(3,10),A(5,12),B(2,9)`
- `P(6,17),A(1,-3),B(3,5)`
- `P(12,23),A(2,8),B(6,14)`
- `P(3,10),A(5,12),B(2,9)`
- `P(6,17),A(1,-3),B(3,5)`
- `P(12,23),A(2,8),B(6,14)`
- `P(17/5,47/5),A(5,13),B(1,4)`
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2. Write down the co-ordinates of the point P that divides the line joining A(-4,1) and B(17,10) in the ratio 1:2
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ratio =
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- `A(5,13),B(1,4),m:n=2:3`
- `A(-4,1),B(17,10),m:n=1:2`
- `A(5,12),B(2,9),m:n=2:1`
- `A(2,8),B(6,14),m:n=5:3` Externally
- `A(1,-3),B(3,5),m:n=5:3` Externally
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3. In what ratio does the x-axis divide the join of `A(2,-3)` and `B(5,6)`? Also find the coordinates of the point of intersection.
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divided by
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- `A(2,-3),B(5,6)` divided by x-axis
- `A(1,2),B(2,3)` divided by x-axis
- `A(5,-6),B(-1,-4)` divided by y-axis
- `A(-2,1),B(4,5)` divided by y-axis
- `A(2,1),B(7,6)` divided by x-axis
- `A(2,-4),B(-3,6)` divided by y-axis
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4. Find the ratio in which the point `P(x,4)` divides the line segment joining the points `A(2,1)` and `B(7,6)`? Also find the value of `x`.
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- `P(x,2),A(12,5),B(4,-3)`
- `P(11,y),A(15,5),B(9,20)`
- `P(-3,y),A(-5,-4),B(-2,3)`
- `P(-4,y),A(-6,10),B(3,-8)`
- `P(x,4),A(2,1),B(7,6)`
- `P(x,0),A(2,-4),B(-3,6)`
- `P(0,y),A(2,-4),B(-3,6)`
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4. Find Midpoint or Trisection points or equidistant points on X-Y axis
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1. Find the coordinates of the midpoint of the line segment joining the points `A(-5,4)` and `B(7,-8)`
2. Find the trisectional points of line joining `A(-3,-5)` and `B(-6,-8)`
3. Find the point on the x-axis which is equidistant from `A(5,4)` and `B(-2,3)`
4. Find the point on the y-axis which is equidistant from `A(6,5)` and `B(-4,3)`
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- `A(-5,4),B(7,-8)`
- `A(2,1),B(1,-3)`
- `A(2,1),B(5,3)`
- `A(3,-5),B(1,1)`
- `A(1,-1),B(-5,-3)`
- `A(-7,-3),B(5,3)`
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5. Find Centroid, Circumcenter, Area of a triangle
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1. Find the centroid of a triangle whose vertices are `A(4,-6),B(3,-2),C(5,2)`
2. Find the circumcentre of a triangle whose vertices are `A(-2,-3),B(-1,0),C(7,-6)`
3. Using determinants, find the area of the triangle with vertices are `A(-3,5),B(3,-6),C(7, 2)`
4. Using determinants show that the following points are collinear `A(2,3),B(-1,-2),C(5,8)`
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- `A(-2,-3),B(-1,0),C(7,-6)`
- `A(-1,0),B(-1,2),C(3,2)`
- `A(-2,3),B(2,-1),C(4,0)`
- `A(1,3),B(-3,5),C(5,-1)`
- `A(1,3),B(0,-2),C(-3,1)`
- `A(1,2),B(3,-4),C(5,-6)`
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6. Find the equation of a line using slope, point, X-intercept, Y-intercept
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1. Find the equation of a straight line passing through `A(-4,5)` and having slope `-2/3`
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Slope :
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- `A(-4,5)`,slope`=-2/3`
- `A(4,5)`,slope`=1`
- `A(-2,3)`,slope`=-4`
- `A(-1,2)`,slope`=-5/4`
- `A(0,3)`,slope`=2`
- `A(0,0)`,slope`=1/4`
- `A(5,4)`,slope`=1/2`
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2. Find the equation of a straight line passing through the points `A(7,5)` and `B(-9,5)`
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- `A(7,5),B(-9,5)`
- `A(-1,1),B(2,-4)`
- `A(-5,-6),B(3,10)`
- `A(3,-5),B(4,-8)`
- `A(-1,-4),B(3,0)`
- `A(7,8),B(1,0)`
- `A(6,4),B(-1,5)`
- `A(2,3),B(7,6)`
- `A(-3,4),B(5,-6)`
- `A(0,7),B(5,-2)`
- `A(0,0),B(-4,-6)`
- `A(3,5),B(6,4)`
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4. Find the slope, x-intercept and y-intercept of the line joining the points `A(1,3)` and `B(3,5)`
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- `A(1,3),B(3,5)`
- `A(4,-8),B(5,-2)`
- `A(7,1),B(8,9)`
- `A(4,8),B(5,5)`
- `A(7,8),B(1,0)`
- `A(6,4),B(-1,5)`
- `A(2,3),B(7,6)`
- `A(-3,4),B(5,-6)`
- `A(0,7),B(5,-2)`
- `A(0,0),B(-4,-6)`
- `A(3,5),B(6,4)`
- `A(3,-5),B(-7,9)`
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8. Find the equation of a line passing through point of intersection of two lines and slope or a point
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1. Find the equation of a line passing through the point of intersection of lines `3x+4y=7` and `x-y+2=0` and having slope 5
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Line-1 : ,
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Line-2 : ,
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Slope :
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- Line-1`:x-4y+18=0`,Line-2`:x+y-12=0`,slope`=2`
- Line-1`:2x+3y+4=0`,Line-2`:3x+6y-8=0`,slope`=2`
- Line-1`:x=3y`,Line-2`:3x=2y+7`,slope`=-1/2`
- Line-1`:x-4y+18=0`,Line-2`:x+y-12=0`,slope`=2`
- Line-1`:2x+3y+4=0`,Line-2`:3x+6y-8=0`,slope`=2`
- Line-1`:x=3y`,Line-2`:3x=2y+7`,slope`=-1/2`
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2. Find the equation of a line passing through the point of intersection of lines `4x+5y+7=0` and `3x-2y-12=0` and point `A(3,1)`
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Line-1 : ,
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Line-2 : ,
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- Line-1`:x+y+1=0`,Line-2`:3x+y-5=0`,`A(1,-3)`
- Line-1`:4x+5y+7=0`,Line-2`:3x-2y-12=0`,`A(3,1)`
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9. Find the equation of a line passing through a point and parallel or perpendicular to Line-2
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1. Find the equation of the line passing through the point `A(5,4)` and parallel to the line `2x+3y+7=0`
2. Find the equation of the line passing through the point `A(1,1)` and perpendicular to the line `2x-3y+2=0`
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Line-2 :
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- `A(5,4)`,Line`:2x+3y+7=0`
- `A(1,1)`,Line`:2x-3y+2=0`
- `A(2,3)`,Line`:2x-3y+8=0`
- `A(2,-5)`,Line`:2x-3y-7=0`
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3. Find the equation of the line passing through the point `A(1,3)` and parallel to line passing through the points `B(3,-5)` and `C(-6,1)`
4. Find the equation of the line passing through the point `A(5,5)` and perpendicular to the line passing through the points `B(1,-2)` and `C(-5,2)`
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- `A(1,3),B(3,-5),C(-6,1)`
- `A(4,-5),B(3,7),C(-2,4)`
- `A(-1,3),B(0,2),C(4,5)`
- `A(2,-3),B(1,2),C(-1,5)`
- `A(4,2),B(1,-1),C(3,2)`
- `A(5,5),B(1,-2),C(-5,2)`
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12. Reflection of points about x-axis, y-axis, origin
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Find Reflection of points A(0,0),B(2,2),C(0,4),D(-2,2) and Reflection about X,Y,O
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Reflection about
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- `A(-2,-2),B(-1,2),C(3,1)` and Reflection about x
- `A(2,3),B(7,4),C(8,7),D(3,6)` and Reflection about y
- `A(1,-1),B(-2,2),C(4,8),D(7,5)` and Reflection about o
- `A(3,0),B(4,5),C(-1,4),D(-2,-1)` and Reflection about x,y
- `A(3,2),B(5,4),C(3,6),D(1,4)` and Reflection about y,x
- `A(-1,-1),B(1,5),C(2,8)` and Reflection about y=x
- `A(-3,2),B(1,2),C(-3,5)` and Reflection about y=-x
- `A(0,-1),B(3,5),C(5,9)` and Reflection about x=2
- `A(2,8),B(1,5),C(0,2)` and Reflection about y=2
- `A(0,0),B(2,2),C(0,4),D(-2,2)` and Reflection about x+3y-7=0
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Solution
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Solution provided by AtoZmath.com
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Circumcenter of a triangle calculator
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1. Find the circumcentre of a triangle whose vertices are `A(-2,-3),B(-1,0),C(7,-6)`
2. Find the circumcentre of a triangle whose vertices are `A(-1,0),B(-1,2),C(3,2)`
3. Find the circumcentre of a triangle whose vertices are `A(-2,3),B(2,-1),C(4,0)`
4. Find the circumcentre of a triangle whose vertices are `A(1,3),B(-3,5),C(5,-1)`
5. Find the circumcentre of a triangle whose vertices are `A(1,3),B(0,-2),C(-3,1)`
6. Find the circumcentre of a triangle whose vertices are `A(1,2),B(3,-4),C(5,-6)`
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Example1. 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`
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