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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
1. Distance, Slope of two points
(Previous method)		2. Determine if the points A(1,5), B(2,3), C(-2,-11) are collinear points
(Next example)

1. A(-1,-1), B(1,5), C(2,8), Find points are collinear points or triangle


1. `A(-1,-1), B(1,5), C(2,8)`, Find points are collinear points or 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)`


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 > BC
If points A, B and C are collinear then AB + BC = AC
Here `2sqrt(10)+sqrt(10)=3sqrt(10)`

`:.` A,B,C are collinear points


2. `A(0,0), B(0,3), C(4,0)`, Find points are collinear points or 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


3. `A(2,2), B(-2,4), C(2,6)`, Find points are collinear points or 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, in which any two sides are equal, is called an isosceles triangle

The given points are `A(2,2),B(-2,4),C(2,6)`

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

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

`=sqrt(16+4)`

`=sqrt(20)`

`:. AB=2sqrt(5)`


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

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

`=sqrt(16+4)`

`=sqrt(20)`

`:. BC=2sqrt(5)`


`AC=sqrt((2-2)^2+(6-2)^2)`

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

`=sqrt(0+16)`

`=sqrt(16)`

`:. AC=4`


Here `AB=BC`

`:.` ABC is an isoceles triangle


4. `A(0,0), B(2,0), C(-4,0), D(-2,0)`, Find points are collinear points or quadrilateral

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)`


The given points are `A(0,0),B(2,0),C(-4,0),D(-2,0)`

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

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

`=sqrt(4+0)`

`=sqrt(4)`

`:. CD=2`


`DA=sqrt((0+2)^2+(0-0)^2)`

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

`=sqrt(4+0)`

`=sqrt(4)`

`:. DA=2`


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

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

`=sqrt(4+0)`

`=sqrt(4)`

`:. AB=2`


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

`=sqrt((6)^2+(0)^2)`

`=sqrt(36+0)`

`=sqrt(36)`

`:. CB=6`


Here `CD+DA+AB=2+2+2=6=CB`

`:.` C,D,A,B are collinear points


5. `A(3,2), B(5,4), C(3,6), D(1,4)`, Find points are collinear points or quadrilateral

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 quadrilateral, in which all sides are equal and also the diagonals are equal, is a square.

So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC=BD`


The given points are `A(3,2),B(5,4),C(3,6),D(1,4)`

Length of sides:

`AB=sqrt((5-3)^2+(4-2)^2)`

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

`=sqrt(4+4)`

`=sqrt(8)`

`:. AB=2sqrt(2)`


`BC=sqrt((3-5)^2+(6-4)^2)`

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

`=sqrt(4+4)`

`=sqrt(8)`

`:. BC=2sqrt(2)`


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

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

`=sqrt(4+4)`

`=sqrt(8)`

`:. CD=2sqrt(2)`


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

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

`=sqrt(4+4)`

`=sqrt(8)`

`:. AD=2sqrt(2)`


Length of diagonals:

`AC=sqrt((3-3)^2+(6-2)^2)`

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

`=sqrt(0+16)`

`=sqrt(16)`

`:. AC=4`


`BD=sqrt((1-5)^2+(4-4)^2)`

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

`=sqrt(16+0)`

`=sqrt(16)`

`:. BD=4`


Here, all sides `AB=BC=CD=AD`

and both diagonals `AC=BD`

Since, all the sides are equal and both the diagonals are equal
Hence, ABCD is a square

6. `A(3,0), B(4,5), C(-1,4), D(-2,-1)`, Find points are collinear points or quadrilateral

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 quadrilateral, in which all sides are equal, is a rhombus.

So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC!=BD`


The given points are `A(3,0),B(4,5),C(-1,4),D(-2,-1)`

Length of sides:

`AB=sqrt((4-3)^2+(5-0)^2)`

`=sqrt((1)^2+(5)^2)`

`=sqrt(1+25)`

`=sqrt(26)`

`:. AB=sqrt(26)`


`BC=sqrt((-1-4)^2+(4-5)^2)`

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

`=sqrt(25+1)`

`=sqrt(26)`

`:. BC=sqrt(26)`


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

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

`=sqrt(1+25)`

`=sqrt(26)`

`:. CD=sqrt(26)`


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

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

`=sqrt(25+1)`

`=sqrt(26)`

`:. AD=sqrt(26)`


Length of diagonals:

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

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

`=sqrt(16+16)`

`=sqrt(32)`

`:. AC=4sqrt(2)`


`BD=sqrt((-2-4)^2+(-1-5)^2)`

`=sqrt((-6)^2+(-6)^2)`

`=sqrt(36+36)`

`=sqrt(72)`

`:. BD=6sqrt(2)`


Here all sides `AB=BC=CD=AD` and both diagonals `AC!=BD`

Since, all the sides are equal and both the diagonals are not equal
Hence, ABCD is a rhombus

Area `=1/2 xx AC xx BD`

`=1/2 xx 4sqrt(2) xx 6sqrt(2)`

`=24`

Hence, the area of the rhombus is `24` square units




This material is intended as a summary. Use your textbook for detail explanation.
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1. Distance, Slope of two points
(Previous method)		2. Determine if the points A(1,5), B(2,3), C(-2,-11) are collinear points
(Next example)


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