Question

show that the following points taken in order from an equilateral triangle in each case A (6,-4),B (-2,-4),C (2,10)

Answer 1

Consider the following points,

A(6,-4) , B(-2,-4) and C(2,10)

Let us consider the distance AB,

For the points $(x_{1},y_{1}) (x_{2},y_{2})$, distance is calculated by distance formula which states,

$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

AB = $\sqrt{(-2-6)^{2}+(-4+4)^{2}}$

AB = 8 units

Now, consider the distance BC

BC = $\sqrt{(-2-2)^{2}+(-4-10)^{2}}$

BC= $\sqrt{212}$ = 14.6 units

Now, consider the distance AC,

AC = $\sqrt{(2-6)^{2}+(-4-10)^{2}}$

AC = $\sqrt{212}$

AC = 14.6 units

Since, the distance of three segments are not equal.

Therefore, it does'not form an equilateral triangle.