Question

if A 4,4 , B 3,-16 , C 3,-2 are the vertices of triangle abc and def are the midpoints of sides BC CA AB respectively then prove that area of triangle abc is 4 times area of triangle def

Answer 1

Step-by-step explanation:

$Let A = (x_1,y_1), B = (x_2,y_2)\ and\ C = (x_3,y_3)$

We are given the coordinates of vertices as:

$(x_1,y_1) = (4,4), (x_2,y_2) = (3,-16) and (x_3,y_3) = (3,-2)$

Since, we know the formula as:  

Area of  triangle ABC $= [x_1 (y_2 - y_3 )+x_2 ( y_3- y_1) + x_3 (y_1 - y_2)]$

Area of triangle ABC =  [4(-16+2) + 3(-2-4) + 3(4+16)]

Area of triangle ABC =  [-56-18+60] = -14 sq. Units.

since, this is area. so it must be positive. so 14 sq units.

Coordinates of d, e and f are  

i.e. $d(\frac {7}{2}, \frac {-12}{2}),e(\frac {6}{2}, \frac {-18}{2}) and f(\frac {7}{2}, \frac {2}{2})$

=$d(\frac{7}{2},-6),e(3,-9),f (\frac{7}{2},1])$

Again we use the formula: Area of  triangle ABC = $[x_1 (y_2 - y_3 ) + x_2 ( y_3- y_1) + x_3 (y_1 - y_2)]$

So, Area of triangle def =$[ \frac {7}{2}(-9-1)+ 3(1+6) + \frac {7}{2}(-6+9)]$

Area of triangle DEF =$[-32 + \frac {63}{2}]=\frac{-7}{2}$.sq units

This is is area. so it can't be negative. so, $\frac{7}{2}$ sq units

(Area of triangle ABC) $= \dfrac{14}{4} =\dfrac{7}{2}=$ Area of triangle DEF

Thus,

Area of triangle ABC = 4  X  Area of triangle DEF