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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 an isosceles triangle calculator
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 1. Show that the points `A(7,10), B(-2,5), C(3,-4)` are vertices of an isosceles triangle
2. Show that the points `A(5,-2), B(6,4), C(7,-2)` are vertices of an isosceles triangle
3. Show that the points `A(3,0), B(6,4), C(-1,3)` are vertices of an isosceles triangle
4. Show that the points `A(-1,4), B(-3,-6), C(3,-2)` are vertices of an isosceles triangle
5. Show that the points `A(2,2), B(-2,4), C(2,6)` are vertices of an isosceles triangle
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Example1. Show that the points `A(7,10), B(-2,5), C(3,-4)` are vertices of an isosceles 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, in which any two sides are equal, is called an isosceles triangle The given points are `A(7,10),B(-2,5),C(3,-4)` `AB=sqrt((-2-7)^2+(5-10)^2)` `=sqrt((-9)^2+(-5)^2)` `=sqrt(81+25)` `=sqrt(106)` `:. AB=sqrt(106)` `BC=sqrt((3+2)^2+(-4-5)^2)` `=sqrt((5)^2+(-9)^2)` `=sqrt(25+81)` `=sqrt(106)` `:. BC=sqrt(106)` `AC=sqrt((3-7)^2+(-4-10)^2)` `=sqrt((-4)^2+(-14)^2)` `=sqrt(16+196)` `=sqrt(212)` `:. AC=2sqrt(53)` Here `AB=BC` `:.` ABC is an isoceles triangle Also `AB^2+BC^2=(sqrt(106))^2+(sqrt(106))^2=106+106=212` and `AC^2=(2sqrt(53))^2=212` `:. AB^2+BC^2=AC^2` and `/_B=90^circ` `:.` ABC is a right angle triangle Hence, ABC is an isoceles right angle triangle 
2. Show that the points `A(5,-2), B(6,4), C(7,-2)` are vertices of an isosceles 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, in which any two sides are equal, is called an isosceles triangle The given points are `A(5,-2),B(6,4),C(7,-2)` `AB=sqrt((6-5)^2+(4+2)^2)` `=sqrt((1)^2+(6)^2)` `=sqrt(1+36)` `=sqrt(37)` `:. AB=sqrt(37)` `BC=sqrt((7-6)^2+(-2-4)^2)` `=sqrt((1)^2+(-6)^2)` `=sqrt(1+36)` `=sqrt(37)` `:. BC=sqrt(37)` `AC=sqrt((7-5)^2+(-2+2)^2)` `=sqrt((2)^2+(0)^2)` `=sqrt(4+0)` `=sqrt(4)` `:. AC=2` Here `AB=BC` `:.` ABC is an isoceles triangle 
3. Show that the points `A(3,0), B(6,4), C(-1,3)` are vertices of an isosceles 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, in which any two sides are equal, is called an isosceles 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=AC` `:.` ABC is an isoceles triangle Also `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 a right angle triangle Hence, ABC is an isoceles right angle triangle 
4. Show that the points `A(-1,4), B(-3,-6), C(3,-2)` are vertices of an isosceles 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, in which any two sides are equal, is called an isosceles triangle The given points are `A(-1,4),B(-3,-6),C(3,-2)` `AB=sqrt((-3+1)^2+(-6-4)^2)` `=sqrt((-2)^2+(-10)^2)` `=sqrt(4+100)` `=sqrt(104)` `:. AB=2sqrt(26)` `BC=sqrt((3+3)^2+(-2+6)^2)` `=sqrt((6)^2+(4)^2)` `=sqrt(36+16)` `=sqrt(52)` `:. BC=2sqrt(13)` `AC=sqrt((3+1)^2+(-2-4)^2)` `=sqrt((4)^2+(-6)^2)` `=sqrt(16+36)` `=sqrt(52)` `:. AC=2sqrt(13)` Here `BC=AC` `:.` ABC is an isoceles triangle Also `BC^2+AC^2=(2sqrt(13))^2+(2sqrt(13))^2=52+52=104` and `AB^2=(2sqrt(26))^2=104` `:. BC^2+AC^2=AB^2` and `/_C=90^circ` `:.` ABC is a right angle triangle Hence, ABC is an isoceles right angle triangle 
5. Show that the points `A(2,2), B(-2,4), C(2,6)` are vertices of an isosceles 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, 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  |
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