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2. Points are Collinear or Triangle or Quadrilateral form example
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- A(-1,-1), B(1,5), C(2,8), Find points are collinear points or triangle
- Determine if the points A(1,5), B(2,3), C(-2,-11) are collinear points
- Show that the points A(-3,0), B(1,-3), C(4,1) are vertices of a right angle triangle
- Show that the points A(1,1), B(-1,-1), C(-1.732051,1.732051) are vertices of an equilateral triangle
- Show that the points A(7,10), B(-2,5), C(3,-4) are vertices of an isosceles triangle
- Determine if the points A(0,0), B(2,0), C(-4,0), D(-2,0) are collinear points
- Show that the points A(1,2), B(5,4), C(3,8), D(-1,6) are vertices of a square
- Show that the points A(-4,-1), B(-2,-4), C(4,0), D(2,3) are vertices of a rectangle
- Show that the points A(3,0), B(4,5), C(-1,4), D(-2,-1) are vertices of a rhombus
- Show that the points A(-3,-2), B(5,-2), C(9,3), D(1,3) are vertices of a parallelogram
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Other related methods
- Distance, Slope of two points
- Points are Collinear or Triangle or Quadrilateral form
- Find Ratio of line joining AB and is divided by P
- Find Midpoint or Trisection points or equidistant points on X-Y axis
- Find Centroid, Circumcenter, Area of a triangle
- Find the equation of a line using slope, point, X-intercept, Y-intercept
- Find Slope, X-intercept, Y-intercept of a line
- Find the equation of a line passing through point of intersection of two lines and slope or a point
- Find the equation of a line passing through a point and parallel or perpendicular to Line-2 or point-2 and point-3
- Find the equation of a line passing through point of intersection of Line-1, Line-2 and parallel or perpendicular to Line-3
- For two lines, find Angle, intersection point and determine if parallel or perpendicular lines
- Reflection of points about x-axis, y-axis, origin
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1. A(-1,-1), B(1,5), C(2,8), Find points are collinear points or triangle
(Previous example) | 3. Show that the points A(-3,0), B(1,-3), C(4,1) are vertices of a right angle triangle
(Next example) |
2. Determine if the points A(1,5), B(2,3), C(-2,-11) are collinear points
1. Determine if the points A(1,5), B(2,3), C(-2,-11) are collinear points
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(1,5),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 > BC
If points A, B and C are collinear then AB + BC = AC
But 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 points
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,-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 > AC
If points A, B and C are collinear then AB + AC = BC
Here 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 points
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
4. Determine if the points A(0,-1), B(3,5), C(5,9) are collinear points
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,-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 > BC
If points A, B and C are collinear then AB + BC = AC
Here 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 points
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(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 > BC
If points A, B and C are collinear then AB + BC = AC
Here sqrt(10)+sqrt(10)=2sqrt(10)
:. A,B,C are collinear points
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
1. A(-1,-1), B(1,5), C(2,8), Find points are collinear points or triangle
(Previous example) | 3. Show that the points A(-3,0), B(1,-3), C(4,1) are vertices of a right angle triangle
(Next example) |
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