Question

Triangle ABC has vertices A(2,0), B(6,0) and C(4,5). Prove that the triangle is isosceles and ​not​ a right triangle

Answer 1

Answer:

Explanation:

An isosceles triangle is one with two equal lengths.

To find the lengths of the sides with coordinates

(

x

1

,

y

1

)

,

(

x

2

,

y

2

)

we use Pythagoras' theorem

d

=

√

(

x

2

−

x

2

)

2

+

(

y

2

−

y

1

)

2

from the information given

d

1

=

√

(

13

−

9

)

2

+

(

−

2

−

−

8

)

2

d

1

=

√

4

2

+

6

2

=

√

52

−

−

(

1

)

d

2

=

√

(

13

−

5

)

2

+

(

−

2

−

−

2

)

2

d

2

=

√

8

2

+

0

2

=

√

64

=

8

−

−

(

2

)

d

3

=

√

(

9

−

5

)

2

+

(

−

8

−

−

2

)

2

d

3

=

√

4

2

+

6

2

=

√

52

−

−

−

(

3

)

we see

d

1

=

d

3Explanation:

An isosceles triangle is one with two equal lengths.

To find the lengths of the sides with coordinates

(

x

1

,

y

1

)

,

(

x

2

,

y

2

)

we use Pythagoras' theorem

d

=

√

(

x

2

−

x

2

)

2

+

(

y

2

−

y

1

)

2

from the information given

d

1

=

√

(

13

−

9

)

2

+

(

−

2

−

−

8

)

2

d

1

=

√

4

2

+

6

2

=

√

52

−

−

(

1

)

d

2

=

√

(

13

−

5

)

2

+

(

−

2

−

−

2

)

2

d

2

=

√

8

2

+

0

2

=

√

64

=

8

−

−

(

2

)

d

3

=

√

(

9

−

5

)

2

+

(

−

8

−

−

2

)

2

d

3

=

√

4

2

+

6

2

=

√

52

−

−

−

(

3

)

we see

d

1

=

d

3Explanation:

An isosceles triangle is one with two equal lengths.

To find the lengths of the sides with coordinates

(

x

1

,

y

1

)

,

(

x

2

,

y

2

)

we use Pythagoras' theorem

d

=

√

(

x

2

−

x

2

)

2

+

(

y

2

−

y

1

)

2

from the information given

d

1

=

√

(

13

−

9

)

2

+

(

−

2

−

−

8

)

2

d

1

=

√

4

2

+

6

2

=

√

52

−

−

(

1

)

d

2

=

√

(

13

−

5

)

2

+

(

−

2

−

−

2

)

2

d

2

=

√

8

2

+

0

2

=

√

64

=

8

−

−

(

2

)

d

3

=

√

(

9

−

5

)

2

+

(

−

8

−

−

2

)

2

d

3

=

√

4

2

+

6

2

=

√

52

−

−

−

(

3

)

we see

d

1

=

d

3