Answer:

†D=

(X

2

−x

1

)

2

+(y

2

−y

1

)

2

Solution :

Let Point be,

A ( 1 , - 4 )

B ( 2 , - 3 )

C ( 4 , - 7 )

\tt\mapsto AB = \sqrt{ {(2 - 1)}^{2} + {( - 4 + 3)}^{2} }↦AB=

(2−1)

2

+(−4+3)

2

\tt\mapsto AB = \sqrt{ {(1)}^{2} + {( - 1)}^{2} }↦AB=

(1)

2

+(−1)

2

\tt\mapsto AB = \sqrt{ 2} \: units.↦AB=

2

units.

\rule{40}1

\tt \mapsto BC = \sqrt{ {(4 - 2)}^{2} + {( - 7 + 3)}^{2} }↦BC=

(4−2)

2

+(−7+3)

2

\tt \mapsto BC = \sqrt{ {(2)}^{2} + {( - 4)}^{2} }↦BC=

(2)

2

+(−4)

2

\tt \mapsto BC = \sqrt{ 4 + 16 }↦BC=

4+16

\tt \mapsto BC = \sqrt{20 } \: units↦BC=

20

units

\rule{40}1

\tt \mapsto AC = \sqrt{ {(4 - 1)}^{2} + {(7 - 4)}^{2} }↦AC=

(4−1)

2

+(7−4)

2

\tt \mapsto AC = \sqrt{9 + 9 }↦AC=

9+9

\tt \mapsto AC = \sqrt{18 } \: units.↦AC=

18

units.

\rule{100}1

Yes,

AB + AC = BC.

And given point ABC lies with 90° angle.

Verification :

\tt \mapsto H^{2} = P^{2} + B^{2}↦H

2

=P

2

+B

2

\tt \mapsto BC^{2} = AB^{2} + AC^{2}↦BC

2

=AB

2

+AC

2

\tt \mapsto( \sqrt{20})^{2} =( \sqrt{ 2})^{2} + (\sqrt{18 } )^{2}↦(

20

)

2

=(

2

)

2

+(

18

)

2

\tt \mapsto 20 = 2+18↦20=2+18

\tt \mapsto 20= 20.↦20=20.