Class 10

Math

All topics

Coordinate Geometry

You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for $ΔABC$ whose vertices are $A(4,−6),B(3,−2)$ and $C(5,2)$.

$$\therefore$$Co-ordinates of $$D$$$$=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)=\left(\dfrac{3+5}{2},\dfrac{-2+2}{2}\right)=(4,0)$$

Median $$AD$$ divide the $$\triangle ABC$$ into two triangles.

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

Here$$(x_1,y_1)=(4,-6)$$

$$(x_2,y_2)=(3,-2)$$

$$(x_3,y_3)=(4,0)$$

Area of $$\triangle ABD=\dfrac{1}{2}\left[4(-2-0)+3(0+6)+4(-6+2)\right]$$

$$= \dfrac{1}{2}\left[-8+18-16 \right]=\dfrac{-6}{2}=3$$

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

Here$$(x_1,y_1)=(4,-6)$$

$$(x_2,y_2)=(4,0)$$

$$(x_3,y_3)=(5,-2)$$

Area of $$\triangle ADC=\dfrac{1}{2}\left[4(0+2)+4(-2+6)+5(-6-0)\right]$$

$$= \dfrac{1}{2}\left[-8+16-30 \right]=\dfrac{-6}{2}=3$$

$$\therefore $$ Area of $$\triangle ABD$$=Area of $$\triangle ADC$$

Hence it is proved that a median of a triangle divides it into two triangles of equal areas.