Question

For abc shown with vertices at A(-2,6),B(8,-2) and C(-8,-4), shown using coordinate geometry that the segment connecting the midpoint of sides Ac and BC is half the length of side AB.

Answer 1

Answer:

It is proved that AB = 2 × DE.

Step-by-step explanation:

The three vertices of triangle ABC are A(-2,6), B(8,-2) and C(-8,-4).

So, the mid point of AC (say D) has coordinates $(\frac{- 2 - 8}{2},\frac{6 - 4}{2}) = (-5,1)$.

And the mid point of BC (say E) has coordinates $(\frac{8 - 8}{2}, \frac{- 2 - 4}{2}) = (0, - 3)$.

Now, the length of DE will be $\sqrt{(- 5 - 0)^{2} + (1 + 3)^{2}} = \sqrt{41}$ units.

Again, the length of AB will be $\sqrt{(- 2 - 8)^{2} + (6 + 2)^{2}} = 2\sqrt{41}$ units.

So, it is proved that AB = 2 × DE. (Answer)