Question

A(7, -3), B(5, 3) and C(3,1) are the vertices of a $△ABC$ and AD is its median. Prove that the median AD divides $△ABC$ into two triangles of equal area.

Answer 1

The vertices of the triangle are A (7,-3), B ( 5,3) and C (3,1)

Co-ordinates of $D=(\frac{5+3}{2},\frac{3+1}{2})=(4,2)$

For the area of the triangle ADC, let

Let $A(7,-3),D(4,2)andC(3,1)$

$(x_1=7,y_1=-3),(x_2=4,y_2=2),(x_3=3,y_3=1)$

Area of triangle ADC

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

$=\frac{1}{2}\left[7\right(2-1)+4(1+3)+3(-3–2\left)\right]$

$=\frac{1}{2}\left[7\right(1)+4(4)+3(-5\left)\right]$

$=\frac{1}{2}[7+16–15]$

$=\frac{1}{2}\left[8\right]$

= 4 sq.unit

For the area of the triangle ABD, let

Let $A(7,-3),B(5,3)andD(4,2)$

$(x_1=7,y_1=-3),(x_2=5,y_2=3),(x_3=4,y_3=2)$

Area of triangle ABD

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

$=\frac{1}{2}\left[7\right(3–2)+5(2+3)+4(-3–3\left)\right]$

$=\frac{1}{2}\left[7\right(1)+5(5)+4(-6\left)\right]$

$=\frac{1}{2}[7+25–24]$

$=\frac{1}{2}\left[8\right]$

= 4 sq.unit

Thus, Area of triangle ADC = Area of triangle ABD = 4 sq.units

Hence, AD divides ΔABC into two triangles of equal areas