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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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1. Determine if the points A(1,5), B(2,3), C(-2,-11) are collinear points
(Previous example) | 3. Show that the points A(1,1), B(-1,-1), C(-1.732051,1.732051) are vertices of an equilateral triangle
(Next example) |
2. 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.
Any bug, improvement, feedback then
1. Determine if the points A(1,5), B(2,3), C(-2,-11) are collinear points
(Previous example) | 3. 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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