| Home > Geometry calculators > Coordinate Geometry > Collinear points calculator | | Method and examples | | Method 1. Distance, Slope from two points 2. Points (3 or 4) are Collinear or Triangle or Quadrilateral form 3. Find Ratio of line joining AB and is divided by P 4. Midpoint or Trisection points or equidistant points on X-Y axis 5. Find Centroid, Circumcenter, Area of a triangle 6. Find equation of a line using slope, point, X-intercept, Y-intercept 7. Find Slope, X-intercept, Y-intercept of a line 8. Find line passing through intersection point of two lines and slope or a point 9. Find line passing through a point and parallel or perpendicular to Line-2 10. Find line passing through intersection point Line-1,Line-2 parallel to Line-3 11. Find Angle, intersection point and determine if parallel lines 12. Reflection of points about x-axis, y-axis, origin | | | | Collinear points | | 2. Points are Collinear or Triangle or Quadrilateral form | | Show that the points are the vertices of 0. auto decide 1. Collinear points (ABC) 2. right angle triangle (ABC) 3. equilateral triangle (ABC) 4. isosceles triangle (ABC) 5. Collinear points (ABCD) 6. square (ABCD) 7. rectangle (ABCD) 8. rhombus (ABCD) 9. parallelogram (ABCD) | | Find `A(0,0), B(2,2), C(0,4), D(-2,2)` are vertices of a square or not A ( , ) , B ( , ) , C ( , ) , D ( , ) | - `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(0,0),B(0,3),C(4,0)` 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(1,-1),B(-2,2),C(4,8),D(7,5)` 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
| | | Mode = Decimal Fraction | | Decimal Place = 0 1 2 3 4 5 6 7 8 9 10 | | | SolutionMethods | Solution | | Solution provided by AtoZmath.com | | Determine if the points are Collinear points calculator |  1. Determine if the points `A(1,5), B(2,3), C(-2,-11)` are collinear points 2. Determine if the points `A(1,-3), B(2,-5), C(-4,7)` are collinear points 3. Determine if the points `A(-1,-1), B(1,5), C(2,8)` are collinear points 4. Determine if the points `A(0,-1), B(3,5), C(5,9)` are collinear points 5. Determine if the points `A(2,8), B(1,5), C(0,2)` are collinear points | Example1. Determine if the points `A(1,5), B(2,3), C(-2,-11)` are collinear pointsSolution:We know that the distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`The given points are `A(1,5),B(2,3),C(-2,-11)``AB=sqrt((2-1)^2+(3-5)^2)``=sqrt((1)^2+(-2)^2)``=sqrt(1+4)``=sqrt(5)``:. AB=sqrt(5)``BC=sqrt((-2-2)^2+(-11-3)^2)``=sqrt((-4)^2+(-14)^2)``=sqrt(16+196)``=sqrt(212)``:. BC=2sqrt(53)``AC=sqrt((-2-1)^2+(-11-5)^2)``=sqrt((-3)^2+(-16)^2)``=sqrt(9+256)``=sqrt(265)``:. AC=sqrt(265)`As, AC > AB and AC > BCIf points A, B and C are collinear then AB + BC = ACBut `sqrt(5)+2sqrt(53)=16.7963!=sqrt(265)``:.` A,B,C are not collinear points 2. Determine if the points `A(1,-3), B(2,-5), C(-4,7)` are collinear pointsSolution:We know that the distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`The given points are `A(1,-3),B(2,-5),C(-4,7)``AB=sqrt((2-1)^2+(-5+3)^2)``=sqrt((1)^2+(-2)^2)``=sqrt(1+4)``=sqrt(5)``:. AB=sqrt(5)``BC=sqrt((-4-2)^2+(7+5)^2)``=sqrt((-6)^2+(12)^2)``=sqrt(36+144)``=sqrt(180)``:. BC=6sqrt(5)``AC=sqrt((-4-1)^2+(7+3)^2)``=sqrt((-5)^2+(10)^2)``=sqrt(25+100)``=sqrt(125)``:. AC=5sqrt(5)`As, BC > AB and BC > ACIf points A, B and C are collinear then AB + AC = BCHere `sqrt(5)+5sqrt(5)=6sqrt(5)``:.` A,B,C are collinear points 3. Determine if the points `A(-1,-1), B(1,5), C(2,8)` are collinear pointsSolution:We know that the distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`The given points are `A(-1,-1),B(1,5),C(2,8)``AB=sqrt((1+1)^2+(5+1)^2)``=sqrt((2)^2+(6)^2)``=sqrt(4+36)``=sqrt(40)``:. AB=2sqrt(10)``BC=sqrt((2-1)^2+(8-5)^2)``=sqrt((1)^2+(3)^2)``=sqrt(1+9)``=sqrt(10)``:. BC=sqrt(10)``AC=sqrt((2+1)^2+(8+1)^2)``=sqrt((3)^2+(9)^2)``=sqrt(9+81)``=sqrt(90)``:. AC=3sqrt(10)`As, AC > AB and AC > BCIf points A, B and C are collinear then AB + BC = ACHere `2sqrt(10)+sqrt(10)=3sqrt(10)``:.` A,B,C are collinear points 4. Determine if the points `A(0,-1), B(3,5), C(5,9)` are collinear pointsSolution:We know that the distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`The given points are `A(0,-1),B(3,5),C(5,9)``AB=sqrt((3-0)^2+(5+1)^2)``=sqrt((3)^2+(6)^2)``=sqrt(9+36)``=sqrt(45)``:. AB=3sqrt(5)``BC=sqrt((5-3)^2+(9-5)^2)``=sqrt((2)^2+(4)^2)``=sqrt(4+16)``=sqrt(20)``:. BC=2sqrt(5)``AC=sqrt((5-0)^2+(9+1)^2)``=sqrt((5)^2+(10)^2)``=sqrt(25+100)``=sqrt(125)``:. AC=5sqrt(5)`As, AC > AB and AC > BCIf points A, B and C are collinear then AB + BC = ACHere `3sqrt(5)+2sqrt(5)=5sqrt(5)``:.` A,B,C are collinear points 5. Determine if the points `A(2,8), B(1,5), C(0,2)` are collinear pointsSolution:We know that the distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`The given points are `A(2,8),B(1,5),C(0,2)``AB=sqrt((1-2)^2+(5-8)^2)``=sqrt((-1)^2+(-3)^2)``=sqrt(1+9)``=sqrt(10)``:. AB=sqrt(10)``BC=sqrt((0-1)^2+(2-5)^2)``=sqrt((-1)^2+(-3)^2)``=sqrt(1+9)``=sqrt(10)``:. BC=sqrt(10)``AC=sqrt((0-2)^2+(2-8)^2)``=sqrt((-2)^2+(-6)^2)``=sqrt(4+36)``=sqrt(40)``:. AC=2sqrt(10)`As, AC > AB and AC > BCIf points A, B and C are collinear then AB + BC = ACHere `sqrt(10)+sqrt(10)=2sqrt(10)``:.` A,B,C are collinear points | | | |      Share this solution or page with your friends. | | |
1. Determine if the points `A(1,5), B(2,3), C(-2,-11)` are collinear points 
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