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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=8Area 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]=0Here, the area of triangle is 0Hence 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=-5Area 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]=0Here, the area of triangle is 0Hence 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=8Area 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]=0Here, the area of triangle is 0Hence 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=14Area 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]=0Here, the area of triangle is 0Hence 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=11Area 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]=4Here, the area of triangle is not 0Hence 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=9Area 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]=0Here, the area of triangle is 0Hence the points are collinear
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