Question

Find the length of three sides of triangle ABC , If the co-ordinates of vertices are:
A( 2 , -5 ) B (-3 , 1) and C( 7 , -3)​
Plz don't write rubbish, guys!​

Answer 1

GiveN :-

• Coordinates of A = ( 2 , - 5 )
• Coordinates of B = ( - 3 , 1 )
• Coordinates of C = ( 7 , - 3 )

To FinD :-

• Length of sides of the triangle

SolutioN :-

By using distance formula

$\bullet \:\:\:\boxed{\bf\orange{ Distance = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2}}} }$

Distance of AB

$\mapsto \sf AB = \sqrt{ {( - 3 - 2)}^{2} + {(1 + 5)}^{2}} \\ \\\mapsto \sf AB = \sqrt{ {( -5)}^{2} + {(6)}^{2}} \\ \\\mapsto \sf AB = \sqrt{25 + 36} \\ \\ \mapsto \boxed{ \sf AB = \sqrt{61} \: unit}$

Distance of BC

$\mapsto \sf BC = \sqrt{ {( 7 + 3)}^{2} + {( - 3 - 1)}^{2}} \\ \\\mapsto \sf BC = \sqrt{ {( 10)}^{2} + {( - 4)}^{2}} \\ \\\mapsto \sf BC = \sqrt{100 + 16} \\ \\ \mapsto \boxed{\sf BC = \sqrt{116} \: unit}$

Distance of AC

$\mapsto \sf AC = \sqrt{ {( 7 - 2)}^{2}+{( - 3+ 5)}^{2}} \\ \\\mapsto \sf AC = \sqrt{ {(5)}^{2} + {(2)}^{2}} \\ \\\mapsto \sf AC = \sqrt{25 +4} \\ \\\mapsto \boxed{\sf AC = \sqrt{29} \: unit}$