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Method and examples
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Method
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Collinear points using determinants
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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,-2),C(5,8)`
- `A(8,1),B(3,-4),C(2,-5)`
- `A(3,-2),B(5,2),C(8,8)`
- `A(3,8),B(-4,2),C(10,14)`
- `A(1,5),B(2,3),C(2,11)`
- `A(2,5),B(5,7),C(8,9)`
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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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Mode =
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Decimal Place =
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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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