Question

area of the triangle formed by joining the mid-points of the sia
angle ABC, whose vertices are A(0, -1), B(2, 1) and C(0, 3).
Aiar​

Answer 1

Answer:

find area then multiply by 1/4 because it is one /4 part of triangle

Step-by-step explanation:

answer is 1 unit

Answer 2

Step-by-step explanation:

Let P be mid point of AB

Q be mid point of BC

R be mid point of AC

Let A(0,-1)=(x1,y1)

B(2,1)=(x2,y2)

C(0,3)=(x3,y3)

Ar(∆ABC)= ½[x1(y2-y3)+X2(y3-y1)+X3(y1-y2)]

=½[0(1-3)+2(3-(-1))+0(-1-1)]

=½[0+2(4)+0]

=½[8]

=4 square units

Area of the triangle ABC is 4 unit square.

We no that the area triangle formed by joining the midpoints of the sides of triangle is ¼ of the area of the whole triangle.

Ar(∆PQR) = ¼ Ar(∆ PQP)

= ¼(4)

= 1 square unit