F.26.Show that the points (12,8),(-2,6) and (6,0 ) are the vertices of aisosceles triangle About the author Aaliyah
Step-by-step explanation: Given:– The points (12,8),(-2,6) and (6,0 ) To find:– Show that the points (12,8),(-2,6) and (6,0 ) are the vertices of an isosceles triangle. Solution:– Given points are (12,8),(-2,6) and (6,0 ) Let A = (12,8) B = (-2,6) C= (6,0) To show that the points A,B,C are the vertices of an Isosceles triangle ABC then we have to show that the lengths of any two sides are equal. Length of AB:– Let (x1, y1)=A(12,8) => x1=12 and y1=8 Let (x2, y2)=B(-2,6)=>x2=-2 and y2=6 We know that Distance formula:- The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units => AB = √[(-2-12)^2+(6+8)^2] => AB = √[(-14)^2+(14)^2] => AB = √(196+196) => AB =√(2×196) AB = 14√2 units ———–(1) Length of BC:– Let (x1, y1)=B(-2,6) => x1=-2 and y1=6 Let (x2, y2)=C(6,0)=>x2=6and y2=0 We know that Distance formula:- The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units => BV = √[(6-(-2))^2+(0-6)^2] => BC = √[(6+2)^2+(-6)^2] => BC = √(64+36) => BC =√(100) BC = 10 units ———–(2) Length of AC:– Let (x1, y1)=A(12,8) => x1=12 and y1=8 Let (x2, y2)=C(6,0)=>x2=6and y2=0 We know that Distance formula:- The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units => AC = √[(6-12)^2+(0-8)^2] => AC = √[(-6)^2+(-8)^2] => AC = √(36+64) => AC =√(100) AC = 10 units ———–(3) From (2)&(3) Length of BC = Length of AC => BC = AC => Lengths of the two sides of the triangle are equal. => A,B,C are the vertices of the Isosceles triangle ABC. Answer:– Given points (12,8),(-2,6) and (6,0 ) are the vertices of the Isosceles triangle. Used formulae:– 1.The lengths of any two sides of a triangle are equal then it is an Isosceles triangle. 2.The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units Reply
Step-by-step explanation:
Given:–
The points (12,8),(-2,6) and (6,0 )
To find:–
Show that the points (12,8),(-2,6) and (6,0 ) are the vertices of an isosceles triangle.
Solution:–
Given points are (12,8),(-2,6) and (6,0 )
Let A = (12,8)
B = (-2,6)
C= (6,0)
To show that the points A,B,C are the vertices of an Isosceles triangle ABC then we have to show that the lengths of any two sides are equal.
Length of AB:–
Let (x1, y1)=A(12,8) => x1=12 and y1=8
Let (x2, y2)=B(-2,6)=>x2=-2 and y2=6
We know that
Distance formula:-
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units
=> AB = √[(-2-12)^2+(6+8)^2]
=> AB = √[(-14)^2+(14)^2]
=> AB = √(196+196)
=> AB =√(2×196)
AB = 14√2 units ———–(1)
Length of BC:–
Let (x1, y1)=B(-2,6) => x1=-2 and y1=6
Let (x2, y2)=C(6,0)=>x2=6and y2=0
We know that
Distance formula:-
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units
=> BV = √[(6-(-2))^2+(0-6)^2]
=> BC = √[(6+2)^2+(-6)^2]
=> BC = √(64+36)
=> BC =√(100)
BC = 10 units ———–(2)
Length of AC:–
Let (x1, y1)=A(12,8) => x1=12 and y1=8
Let (x2, y2)=C(6,0)=>x2=6and y2=0
We know that
Distance formula:-
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units
=> AC = √[(6-12)^2+(0-8)^2]
=> AC = √[(-6)^2+(-8)^2]
=> AC = √(36+64)
=> AC =√(100)
AC = 10 units ———–(3)
From (2)&(3)
Length of BC = Length of AC
=> BC = AC
=> Lengths of the two sides of the triangle are equal.
=> A,B,C are the vertices of the Isosceles triangle ABC.
Answer:–
Given points (12,8),(-2,6) and (6,0 ) are the vertices of the Isosceles triangle.
Used formulae:–
1.The lengths of any two sides of a triangle are equal then it is an Isosceles triangle.
2.The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units