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2. Points are Collinear or Triangle or Quadrilateral form example
( Enter your problem )
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- 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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9. Show that the points A(-3,-2), B(5,-2), C(9,3), D(1,3) are vertices of a parallelogram
1. Show that the points `A(-3,-2), B(5,-2), C(9,3), D(1,3)` are vertices of a parallelogram
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 opposites sides are equal, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD`
The given points are `A(-3,-2),B(5,-2),C(9,3),D(1,3)`
Length of sides:
`AB=sqrt((5+3)^2+(-2+2)^2)`
`=sqrt((8)^2+(0)^2)`
`=sqrt(64+0)`
`=sqrt(64)`
`:. AB=8`
`CD=sqrt((1-9)^2+(3-3)^2)`
`=sqrt((-8)^2+(0)^2)`
`=sqrt(64+0)`
`=sqrt(64)`
`:. CD=8`
`BC=sqrt((9-5)^2+(3+2)^2)`
`=sqrt((4)^2+(5)^2)`
`=sqrt(16+25)`
`=sqrt(41)`
`:. BC=sqrt(41)`
`AD=sqrt((1+3)^2+(3+2)^2)`
`=sqrt((4)^2+(5)^2)`
`=sqrt(16+25)`
`=sqrt(41)`
`:. AD=sqrt(41)`
Here, opposite sides `AB=CD` and `BC=AD`
Since, the lengths of the opposite sides are equal
Hence, ABCD is a parallelogram
2. Show that the points `A(7,3), B(6,1), C(8,2), D(9,4)` are vertices of a parallelogram
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 opposites sides are equal, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD`
The given points are `A(7,3),B(6,1),C(8,2),D(9,4)`
Length of sides:
`AB=sqrt((6-7)^2+(1-3)^2)`
`=sqrt((-1)^2+(-2)^2)`
`=sqrt(1+4)`
`=sqrt(5)`
`:. AB=sqrt(5)`
`CD=sqrt((9-8)^2+(4-2)^2)`
`=sqrt((1)^2+(2)^2)`
`=sqrt(1+4)`
`=sqrt(5)`
`:. CD=sqrt(5)`
`BC=sqrt((8-6)^2+(2-1)^2)`
`=sqrt((2)^2+(1)^2)`
`=sqrt(4+1)`
`=sqrt(5)`
`:. BC=sqrt(5)`
`AD=sqrt((9-7)^2+(4-3)^2)`
`=sqrt((2)^2+(1)^2)`
`=sqrt(4+1)`
`=sqrt(5)`
`:. AD=sqrt(5)`
Here, opposite sides `AB=CD` and `BC=AD`
Since, the lengths of the opposite sides are equal
Hence, ABCD is a parallelogram
3. Show that the points `A(1,-2), B(3,6), C(5,10), D(3,2)` are vertices of a parallelogram
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 opposites sides are equal, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD`
The given points are `A(1,-2),B(3,6),C(5,10),D(3,2)`
Length of sides:
`AB=sqrt((3-1)^2+(6+2)^2)`
`=sqrt((2)^2+(8)^2)`
`=sqrt(4+64)`
`=sqrt(68)`
`:. AB=2sqrt(17)`
`CD=sqrt((3-5)^2+(2-10)^2)`
`=sqrt((-2)^2+(-8)^2)`
`=sqrt(4+64)`
`=sqrt(68)`
`:. CD=2sqrt(17)`
`BC=sqrt((5-3)^2+(10-6)^2)`
`=sqrt((2)^2+(4)^2)`
`=sqrt(4+16)`
`=sqrt(20)`
`:. BC=2sqrt(5)`
`AD=sqrt((3-1)^2+(2+2)^2)`
`=sqrt((2)^2+(4)^2)`
`=sqrt(4+16)`
`=sqrt(20)`
`:. AD=2sqrt(5)`
Here, opposite sides `AB=CD` and `BC=AD`
Since, the lengths of the opposite sides are equal
Hence, ABCD is a parallelogram
4. Show that the points `A(6,8), B(3,7), C(-2,-2), D(1,-1)` are vertices of a parallelogram
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 opposites sides are equal, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD`
The given points are `A(6,8),B(3,7),C(-2,-2),D(1,-1)`
Length of sides:
`AB=sqrt((3-6)^2+(7-8)^2)`
`=sqrt((-3)^2+(-1)^2)`
`=sqrt(9+1)`
`=sqrt(10)`
`:. AB=sqrt(10)`
`CD=sqrt((1+2)^2+(-1+2)^2)`
`=sqrt((3)^2+(1)^2)`
`=sqrt(9+1)`
`=sqrt(10)`
`:. CD=sqrt(10)`
`BC=sqrt((-2-3)^2+(-2-7)^2)`
`=sqrt((-5)^2+(-9)^2)`
`=sqrt(25+81)`
`=sqrt(106)`
`:. BC=sqrt(106)`
`AD=sqrt((1-6)^2+(-1-8)^2)`
`=sqrt((-5)^2+(-9)^2)`
`=sqrt(25+81)`
`=sqrt(106)`
`:. AD=sqrt(106)`
Here, opposite sides `AB=CD` and `BC=AD`
Since, the lengths of the opposite sides are equal
Hence, ABCD is a parallelogram
5. Show that the points `A(3,1), B(0,-2), C(1,1), D(4,4)` are vertices of a parallelogram
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 opposites sides are equal, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD`
The given points are `A(3,1),B(0,-2),C(1,1),D(4,4)`
Length of sides:
`AB=sqrt((0-3)^2+(-2-1)^2)`
`=sqrt((-3)^2+(-3)^2)`
`=sqrt(9+9)`
`=sqrt(18)`
`:. AB=3sqrt(2)`
`CD=sqrt((4-1)^2+(4-1)^2)`
`=sqrt((3)^2+(3)^2)`
`=sqrt(9+9)`
`=sqrt(18)`
`:. CD=3sqrt(2)`
`BC=sqrt((1-0)^2+(1+2)^2)`
`=sqrt((1)^2+(3)^2)`
`=sqrt(1+9)`
`=sqrt(10)`
`:. BC=sqrt(10)`
`AD=sqrt((4-3)^2+(4-1)^2)`
`=sqrt((1)^2+(3)^2)`
`=sqrt(1+9)`
`=sqrt(10)`
`:. AD=sqrt(10)`
Here, opposite sides `AB=CD` and `BC=AD`
Since, the lengths of the opposite sides are equal
Hence, ABCD is a parallelogram
6. Show that the points `A(2,1), B(5,2), C(6,4), D(3,3)` are vertices of a parallelogram
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 opposites sides are equal, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD`
The given points are `A(2,1),B(5,2),C(6,4),D(3,3)`
Length of sides:
`AB=sqrt((5-2)^2+(2-1)^2)`
`=sqrt((3)^2+(1)^2)`
`=sqrt(9+1)`
`=sqrt(10)`
`:. AB=sqrt(10)`
`CD=sqrt((3-6)^2+(3-4)^2)`
`=sqrt((-3)^2+(-1)^2)`
`=sqrt(9+1)`
`=sqrt(10)`
`:. CD=sqrt(10)`
`BC=sqrt((6-5)^2+(4-2)^2)`
`=sqrt((1)^2+(2)^2)`
`=sqrt(1+4)`
`=sqrt(5)`
`:. BC=sqrt(5)`
`AD=sqrt((3-2)^2+(3-1)^2)`
`=sqrt((1)^2+(2)^2)`
`=sqrt(1+4)`
`=sqrt(5)`
`:. AD=sqrt(5)`
Here, opposite sides `AB=CD` and `BC=AD`
Since, the lengths of the opposite sides are equal
Hence, ABCD is a parallelogram
This material is intended as a summary. Use your textbook for detail explanation.
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