Question

F.
26.
Show that the points (12,8),(-2,6) and (6,0 ) are the vertices of a
isosceles triangle
​

Answer 1

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