Question

If A (4,-6), B (3, -2) and C (5, 2) are the vertices of Δ ABC, then verify the fact that a median of a triangle ABC divides it into two triangles of equal areas

Answer 1

Answer:

Step-by-step explanation:

as D is the median on BC so it divide BC in two equal part so, coordinate of D is {(x of B + x of C)/2, (y of B + y of C)/2}  therefore D(4,0)

find area of triangles ADB and ADC  

Area of triangle ADB= 1/2{x1(y2 -y3)+x2(y3 - y1)+x3(y1-y2)}

                                    =1/2{4(-2-0)+3(0-(-6))+4(-6-(-2))}

                                    =1/2{-8+18+-16}

                                    =1/2{-6}

                                    =-3

but, area cannot be negative.

So, area of triangle ADB= 3 square units

Similarly, Area of triangle ADC= 3 square units

therefore, Area of triangle ADB= Area of triangle ADC

Hence, Proved.

therefore ,

Area of triangle ABC= 1/2{4(-2-2)+3(2-(-6))+5(-6-(-2))}

                                   =1/2{-16+24-20}

                                  =-6

But, Area cannot be negative.

therefore, area of triangle ABC= 6 square units

Area of triangle ABC= Area of triangle ADC/2 = Area of triangle ADB/2

Hence, verified.

Answer 2

Step-by-step explanation:

Find the coordinates of the mid-point of base and find areas of both the Triangles separately

Please mark it BRAINLIEST