Question

The ratio of the area of, a triangle ABC with vertices A(0, -1), B(2, 1), C(0, 3) and the triangle formed by joining the mid points of given triangle, is

• A. 1 : 4
• B. 4 : 1
• C. 2 : 3
• D. 1 : 2

Answer 1

The correct option is B 4 : 1

Area of triangle ABC =
$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$=\frac{1}{2}\left[0\right(1-3)+2(3+1)+0(0-1\left)\right]$

$=\frac{1}{2}[0+8+0]$

$=4$

Let the midpoints of AB, BC, and CA be D, E and F respectively

$\therefore D=(\frac{0+2}{2},\frac{-1+1}{2})=(1,0)E=(\frac{2+0}{2},\frac{1+3}{2})=(1,2)F=(\frac{0+0}{2},\frac{3-1}{2})=(0,1)$

Area of triangle DEF =
$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$=\frac{1}{2}\left[1\right(2-1)+1(1-0)+0(1-0\left)\right]$

$=\frac{1}{2}[1+1]$

$=1$
So, the ratio of the area of, a triangle ABC and the triangle formed by joining the mid points of given triangle is 4 : 1