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Solution
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Solution provided by AtoZmath.com
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Collinear points using determinants calculator
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 1. Using determinants show that the following points are collinear `A(2,3),B(-1,-2),C(5,8)`
2. Using determinants show that the following points are collinear `A(8,1),B(3,-4),C(2,-5)`
3. Using determinants show that the following points are collinear `A(3,-2),B(5,2),C(8,8)`
4. Using determinants show that the following points are collinear `A(3,8),B(-4,2),C(10,14)`
5. Using determinants show that the following points are collinear `A(1,5),B(2,3),C(2,11)`
6. Using determinants show that the following points are collinear `A(2,5),B(5,7),C(8,9)`
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Example1. 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 
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