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2. Points are Collinear or Triangle or Quadrilateral form example ( Enter your problem )
( Enter your problem )
  1. A(-1,-1), B(1,5), C(2,8), Find points are collinear points or triangle
  2. Determine if the points A(1,5), B(2,3), C(-2,-11) are collinear points
  3. Show that the points A(-3,0), B(1,-3), C(4,1) are vertices of a right angle triangle
  4. Show that the points A(1,1), B(-1,-1), C(-1.732051,1.732051) are vertices of an equilateral triangle
  5. Show that the points A(7,10), B(-2,5), C(3,-4) are vertices of an isosceles triangle
  6. Determine if the points A(0,0), B(2,0), C(-4,0), D(-2,0) are collinear points
  7. Show that the points A(1,2), B(5,4), C(3,8), D(-1,6) are vertices of a square
  8. Show that the points A(-4,-1), B(-2,-4), C(4,0), D(2,3) are vertices of a rectangle
  9. Show that the points A(3,0), B(4,5), C(-1,4), D(-2,-1) are vertices of a rhombus
  10. Show that the points A(-3,-2), B(5,-2), C(9,3), D(1,3) are vertices of a parallelogram
 		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
2. Determine if the points A(1,5), B(2,3), C(-2,-11) are collinear points
(Previous example)		4. Show that the points A(1,1), B(-1,-1), C(-1.732051,1.732051) are vertices of an equilateral triangle
(Next example)

3. Show that the points A(-3,0), B(1,-3), C(4,1) are vertices of a right angle triangle


1. Show that the points `A(-3,0), B(1,-3), C(4,1)` are vertices of a right angle triangle

Solution:
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)`


A Triangle, which satisfies the Pythagoras theorem, is called a right angled triangle

The given points are `A(-3,0),B(1,-3),C(4,1)`

`AB=sqrt((1+3)^2+(-3-0)^2)`

`=sqrt((4)^2+(-3)^2)`

`=sqrt(16+9)`

`=sqrt(25)`

`:. AB=5`


`BC=sqrt((4-1)^2+(1+3)^2)`

`=sqrt((3)^2+(4)^2)`

`=sqrt(9+16)`

`=sqrt(25)`

`:. BC=5`


`AC=sqrt((4+3)^2+(1-0)^2)`

`=sqrt((7)^2+(1)^2)`

`=sqrt(49+1)`

`=sqrt(50)`

`:. AC=5sqrt(2)`


Here `AB^2+BC^2=(5)^2+(5)^2=25+25=50`

and `AC^2=(5sqrt(2))^2=50`

`:. AB^2+BC^2=AC^2` and `/_B=90^circ`

`:.` ABC is an right angle triangle


Also `AB=BC`

`:.` ABC is an isoceles triangle

Hence, ABC is an isoceles right angle triangle


2. Show that the points `A(3,0), B(6,4), C(-1,3)` are vertices of a right angle triangle

Solution:
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)`


A Triangle, which satisfies the Pythagoras theorem, is called a right angled triangle

The given points are `A(3,0),B(6,4),C(-1,3)`

`AB=sqrt((6-3)^2+(4-0)^2)`

`=sqrt((3)^2+(4)^2)`

`=sqrt(9+16)`

`=sqrt(25)`

`:. AB=5`


`BC=sqrt((-1-6)^2+(3-4)^2)`

`=sqrt((-7)^2+(-1)^2)`

`=sqrt(49+1)`

`=sqrt(50)`

`:. BC=5sqrt(2)`


`AC=sqrt((-1-3)^2+(3-0)^2)`

`=sqrt((-4)^2+(3)^2)`

`=sqrt(16+9)`

`=sqrt(25)`

`:. AC=5`


Here `AB^2+AC^2=(5)^2+(5)^2=25+25=50`

and `BC^2=(5sqrt(2))^2=50`

`:. AB^2+AC^2=BC^2` and `/_A=90^circ`

`:.` ABC is an right angle triangle


Also `AB=AC`

`:.` ABC is an isoceles triangle

Hence, ABC is an isoceles right angle triangle


3. Show that the points `A(0,0), B(0,3), C(4,0)` are vertices of a right angle triangle

Solution:
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)`


A Triangle, which satisfies the Pythagoras theorem, is called a right angled triangle

The given points are `A(0,0),B(0,3),C(4,0)`

`AB=sqrt((0-0)^2+(3-0)^2)`

`=sqrt((0)^2+(3)^2)`

`=sqrt(0+9)`

`=sqrt(9)`

`:. AB=3`


`BC=sqrt((4-0)^2+(0-3)^2)`

`=sqrt((4)^2+(-3)^2)`

`=sqrt(16+9)`

`=sqrt(25)`

`:. BC=5`


`AC=sqrt((4-0)^2+(0-0)^2)`

`=sqrt((4)^2+(0)^2)`

`=sqrt(16+0)`

`=sqrt(16)`

`:. AC=4`


Here `AB^2+AC^2=(3)^2+(4)^2=9+16=25`

and `BC^2=(5)^2=25`

`:. AB^2+AC^2=BC^2` and `/_A=90^circ`

`:.` ABC is an right angle triangle



4. Show that the points `A(-2,-2), B(-1,2), C(3,1)` are vertices of a right angle triangle

Solution:
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)`


A Triangle, which satisfies the Pythagoras theorem, is called a right angled triangle

The given points are `A(-2,-2),B(-1,2),C(3,1)`

`AB=sqrt((-1+2)^2+(2+2)^2)`

`=sqrt((1)^2+(4)^2)`

`=sqrt(1+16)`

`=sqrt(17)`

`:. AB=sqrt(17)`


`BC=sqrt((3+1)^2+(1-2)^2)`

`=sqrt((4)^2+(-1)^2)`

`=sqrt(16+1)`

`=sqrt(17)`

`:. BC=sqrt(17)`


`AC=sqrt((3+2)^2+(1+2)^2)`

`=sqrt((5)^2+(3)^2)`

`=sqrt(25+9)`

`=sqrt(34)`

`:. AC=sqrt(34)`


Here `AB^2+BC^2=(sqrt(17))^2+(sqrt(17))^2=17+17=34`

and `AC^2=(sqrt(34))^2=34`

`:. AB^2+BC^2=AC^2` and `/_B=90^circ`

`:.` ABC is an right angle triangle


Also `AB=BC`

`:.` ABC is an isoceles triangle

Hence, ABC is an isoceles right angle triangle


5. Show that the points `A(-3,2), B(1,2), C(-3,5)` are vertices of a right angle triangle

Solution:
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)`


A Triangle, which satisfies the Pythagoras theorem, is called a right angled triangle

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

`AB=sqrt((1+3)^2+(2-2)^2)`

`=sqrt((4)^2+(0)^2)`

`=sqrt(16+0)`

`=sqrt(16)`

`:. AB=4`


`BC=sqrt((-3-1)^2+(5-2)^2)`

`=sqrt((-4)^2+(3)^2)`

`=sqrt(16+9)`

`=sqrt(25)`

`:. BC=5`


`AC=sqrt((-3+3)^2+(5-2)^2)`

`=sqrt((0)^2+(3)^2)`

`=sqrt(0+9)`

`=sqrt(9)`

`:. AC=3`


Here `AB^2+AC^2=(4)^2+(3)^2=16+9=25`

and `BC^2=(5)^2=25`

`:. AB^2+AC^2=BC^2` and `/_A=90^circ`

`:.` ABC is an right angle triangle





This material is intended as a summary. Use your textbook for detail explanation.
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2. Determine if the points A(1,5), B(2,3), C(-2,-11) are collinear points
(Previous example)		4. Show that the points A(1,1), B(-1,-1), C(-1.732051,1.732051) are vertices of an equilateral triangle
(Next example)


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