5. Find Centroid, Circumcenter, Area of a triangle example ( Enter your problem )
  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)
Other related methods
  1. Distance, Slope of two points
  2. Points are Collinear or Triangle or Quadrilateral form
  3. Find Ratio of line joining AB and is divided by P
  4. Find Midpoint or Trisection points or equidistant points on X-Y axis
  5. Find Centroid, Circumcenter, Area of a triangle
  6. Find the equation of a line using slope, point, X-intercept, Y-intercept
  7. Find Slope, X-intercept, Y-intercept of a line
  8. Find the equation of a line passing through point of intersection of two lines and slope or a point
  9. Find the equation of a line passing through a point and parallel or perpendicular to Line-2 or point-2 and point-3
  10. Find the equation of a line passing through point of intersection of Line-1, Line-2 and parallel or perpendicular to Line-3
  11. For two lines, find Angle, intersection point and determine if parallel or perpendicular lines
  12. Reflection of points about x-axis, y-axis, origin

3. Using determinants, find the area of the triangle with vertices are A(-3,5),B(3,-6),C(7, 2)
(Previous example)
6. Find the equation of a line using slope, point, X-intercept, Y-intercept
(Next method)

4. Using determinants show that the following points are collinear A(2,3),B(-1,-2),C(5,8)





1. Using determinants show that the following points are collinear `A(2,3),B(-1,-2),C(5,8)`

Solution:
The given points are `A(2,3),B(-1,-2),C(5,8)`

`:. x_1=2,y_1=3,x_2=-1,y_2=-2,x_3=5,y_3=8`

Area of a triangle `=1/2 |[x_1,y_1,1],[x_2,y_2,1],[x_2,y_2,1]|`

`=1/2 |[2,3,1],[-1,-2,1],[5,8,1]|`

`=1/2[2 xx (-2 × 1 - 1 × 8) -3 xx (-1 × 1 - 1 × 5) +1 xx (-1 × 8 - (-2) × 5)]`

`=1/2[2 xx (-2 -8) -3 xx (-1 -5) +1 xx (-8 +10)]`

`=1/2[2 xx (-10) -3 xx (-6) +1 xx (2)]`

`=1/2[-20 +18 +2]`

`=1/2[0]`

`=0`

Here, the area of triangle is `0`

Hence the points are collinear



2. Using determinants show that the following points are collinear `A(8,1),B(3,-4),C(2,-5)`

Solution:
The given points are `A(8,1),B(3,-4),C(2,-5)`

`:. x_1=8,y_1=1,x_2=3,y_2=-4,x_3=2,y_3=-5`

Area of a triangle `=1/2 |[x_1,y_1,1],[x_2,y_2,1],[x_2,y_2,1]|`

`=1/2 |[8,1,1],[3,-4,1],[2,-5,1]|`

`=1/2[8 xx (-4 × 1 - 1 × (-5)) -1 xx (3 × 1 - 1 × 2) +1 xx (3 × (-5) - (-4) × 2)]`

`=1/2[8 xx (-4 +5) -1 xx (3 -2) +1 xx (-15 +8)]`

`=1/2[8 xx (1) -1 xx (1) +1 xx (-7)]`

`=1/2[8 -1 -7]`

`=1/2[0]`

`=0`

Here, the area of triangle is `0`

Hence the points are collinear



3. Using determinants show that the following points are collinear `A(3,-2),B(5,2),C(8,8)`

Solution:
The given points are `A(3,-2),B(5,2),C(8,8)`

`:. x_1=3,y_1=-2,x_2=5,y_2=2,x_3=8,y_3=8`

Area of a triangle `=1/2 |[x_1,y_1,1],[x_2,y_2,1],[x_2,y_2,1]|`

`=1/2 |[3,-2,1],[5,2,1],[8,8,1]|`

`=1/2[3 xx (2 × 1 - 1 × 8) +2 xx (5 × 1 - 1 × 8) +1 xx (5 × 8 - 2 × 8)]`

`=1/2[3 xx (2 -8) +2 xx (5 -8) +1 xx (40 -16)]`

`=1/2[3 xx (-6) +2 xx (-3) +1 xx (24)]`

`=1/2[-18 -6 +24]`

`=1/2[0]`

`=0`

Here, the area of triangle is `0`

Hence the points are collinear



4. Using determinants show that the following points are collinear `A(3,8),B(-4,2),C(10,14)`

Solution:
The given points are `A(3,8),B(-4,2),C(10,14)`

`:. x_1=3,y_1=8,x_2=-4,y_2=2,x_3=10,y_3=14`

Area of a triangle `=1/2 |[x_1,y_1,1],[x_2,y_2,1],[x_2,y_2,1]|`

`=1/2 |[3,8,1],[-4,2,1],[10,14,1]|`

`=1/2[3 xx (2 × 1 - 1 × 14) -8 xx (-4 × 1 - 1 × 10) +1 xx (-4 × 14 - 2 × 10)]`

`=1/2[3 xx (2 -14) -8 xx (-4 -10) +1 xx (-56 -20)]`

`=1/2[3 xx (-12) -8 xx (-14) +1 xx (-76)]`

`=1/2[-36 +112 -76]`

`=1/2[0]`

`=0`

Here, the area of triangle is `0`

Hence the points are collinear



5. Using determinants show that the following points are collinear `A(1,5),B(2,3),C(2,11)`

Solution:
The given points are `A(1,5),B(2,3),C(2,11)`

`:. x_1=1,y_1=5,x_2=2,y_2=3,x_3=2,y_3=11`

Area of a triangle `=1/2 |[x_1,y_1,1],[x_2,y_2,1],[x_2,y_2,1]|`

`=1/2 |[1,5,1],[2,3,1],[2,11,1]|`

`=1/2[1 xx (3 × 1 - 1 × 11) -5 xx (2 × 1 - 1 × 2) +1 xx (2 × 11 - 3 × 2)]`

`=1/2[1 xx (3 -11) -5 xx (2 -2) +1 xx (22 -6)]`

`=1/2[1 xx (-8) -5 xx (0) +1 xx (16)]`

`=1/2[-8 +0 +16]`

`=1/2[8]`

`=4`

Here, the area of triangle is not `0`

Hence Given point are not collinear



6. Using determinants show that the following points are collinear `A(2,5),B(5,7),C(8,9)`

Solution:
The given points are `A(2,5),B(5,7),C(8,9)`

`:. x_1=2,y_1=5,x_2=5,y_2=7,x_3=8,y_3=9`

Area of a triangle `=1/2 |[x_1,y_1,1],[x_2,y_2,1],[x_2,y_2,1]|`

`=1/2 |[2,5,1],[5,7,1],[8,9,1]|`

`=1/2[2 xx (7 × 1 - 1 × 9) -5 xx (5 × 1 - 1 × 8) +1 xx (5 × 9 - 7 × 8)]`

`=1/2[2 xx (7 -9) -5 xx (5 -8) +1 xx (45 -56)]`

`=1/2[2 xx (-2) -5 xx (-3) +1 xx (-11)]`

`=1/2[-4 +15 -11]`

`=1/2[0]`

`=0`

Here, the area of triangle is `0`

Hence the points are collinear





This material is intended as a summary. Use your textbook for detail explanation.
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3. Using determinants, find the area of the triangle with vertices are A(-3,5),B(3,-6),C(7, 2)
(Previous example)
6. Find the equation of a line using slope, point, X-intercept, Y-intercept
(Next method)





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