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Solution
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Solution provided by AtoZmath.com
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Show that the points are the vertices of a right angled triangle calculator
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 1. Show that the points `A(-3,0), B(1,-3), C(4,1)` are vertices of a right angle triangle
2. Show that the points `A(3,0), B(6,4), C(-1,3)` are vertices of a right angle triangle
3. Show that the points `A(0,0), B(0,3), C(4,0)` are vertices of a right angle triangle
4. Show that the points `A(-2,-2), B(-1,2), C(3,1)` are vertices of a right angle triangle
5. Show that the points `A(-3,2), B(1,2), C(-3,5)` are vertices of a right angle triangle
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Example1. Show that the points `A(-3,0), B(1,-3), C(4,1)` are vertices of a right angle triangleSolution: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 triangleSolution: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 triangleSolution: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 triangleSolution: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 triangleSolution: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  |
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