Question

without using Pythagoras theorem, show that the points (1,-4)(2,-3) and (4,-7) form right angled triangle.​

Answer 1

Formula Use :

$\tt\dagger \: \: \: \: \: D = \sqrt{ {(x_2 - x_1 )}^{2} + {( y_2 - y_1 )}^{2} }$

Solution :

Let Point be,

• A ( 1 , - 4 )
• B ( 2 , - 3 )
• C ( 4 , - 7 )

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

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

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

$\rule{40}1$

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

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

$\tt \mapsto BC = \sqrt{ 4 + 16 }$

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

$\rule{40}1$

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

$\tt \mapsto AC = \sqrt{9 + 9 }$

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

$\rule{100}1$

Yes,

• AB + AC = BC.
• And given point ABC lies with 90° angle.

Verification :

$\tt \mapsto H^{2} = P^{2} + B^{2}$

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

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

$\tt \mapsto 20 = 2+18$

$\tt \mapsto 20= 20.$