Answer:
Solution given:
A.coordinate are
A(-2,3)
B(0,-3)
C(4,5)
B.Each length are :
we have
length [tex] \sqrt{(x2-x1)²+(y2-y1)²} [/tex]
now
AB:[tex] \sqrt{(-2-0)²+(3+3)²} [/tex]=[tex] 2\sqrt{10} [/tex]units
BC:[tex] \sqrt{(0-4)²+(-3-5)²} [/tex]=[tex] 4\sqrt{5} [/tex]units
AC:[tex] \sqrt{(-2-4)²+(3-5)²} [/tex]=[tex] 2\sqrt{10} [/tex]units
C. the figure:
By using Pythagoras law
base[b]=AB=perpendicular [p]=AC
hypotenuse [h]=BC
we have
h²=p²+b²
substituting value
([tex] 4\sqrt{5} [/tex])²=2p²
16*5=2*([tex] 2\sqrt{10} [/tex])²
80=2*4*10
80=80
SO IT IS RIGHT ANGLED ISOSCELES TRIANGLE.
