Question

A. What are the coordinates of the vertices of the triangle below?

Answer 1

Answer:

Solution given:

A.coordinate are

A(-2,3)

B(0,-3)

C(4,5)

B.Each length are :

we have

length $\sqrt{(x2-x1)²+(y2-y1)²}$

now

AB:$\sqrt{(-2-0)²+(3+3)²}$=$2\sqrt{10}$units

BC:$\sqrt{(0-4)²+(-3-5)²}$=$4\sqrt{5}$units

AC:$\sqrt{(-2-4)²+(3-5)²}$=$2\sqrt{10}$units

C. the figure:

By using Pythagoras law

base[b]=AB=perpendicular [p]=AC

hypotenuse [h]=BC

we have

h²=p²+b²

substituting value

($4\sqrt{5}$)²=2p²

16*5=2*($2\sqrt{10}$)²

80=2*4*10

80=80

SO IT IS RIGHT ANGLED ISOSCELES TRIANGLE.