Question

Find the lengths of the sides of the triangle whose
vertices are A(3, 4), B(2, -1) and C(4, -6).​

Answer 1

Step-by-step explanation:

use distance formula:

√[ (x2-x1)^2+(y2-y1)^2]

AB = √[(2-3)^2+(-1-4)^2] = √26

BC = √[(4-2)^2+(-6-(-1) )^2] = √29

CA =√[ (3-4)^2 + {4-(-6)}^2 ] = √101

Answer 2

Step-by-step explanation:

We Know that,

$\underline{\boxed{\sf Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}$

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

$\sf : \implies \sqrt{(1)^2+(5)^2}$

$\sf : \implies \sqrt{1+25}$

$\sf : \implies \sqrt{26}$

$\sf \red{BC} = \sqrt{(4-2)^2+(6-1)^2}$

$\sf : \implies \sqrt{(2)^2+(5)^2}$

$\sf : \implies \sqrt{4+25}$

$\sf : \implies \sqrt{29}$

$\sf \red{AC} = \sqrt{(4-3)^2+(4+6)^2}$

$\sf : \implies \sqrt{(1)^2+(10)^2}$

$\sf : \implies \sqrt{1+100}$

$\sf : \implies \sqrt{101}$