Question

One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is
(a) (–1, –1)
(b) (2, 2)
(c) (–2, –2)
(d) (2, –2)

Answer 1

Let ABC be the equilateral triangle with vertex A(a, b) and centroid (0, 0)

Since AG ⊥ BC
Slope of BC is −1
⇒ slope of AG is 1
i.e. $\frac{b-0}{a-0}=1$
i.e. b = a
Now distance of (0, 0) from BC

$$\begin{gathered} =\left|\frac{0+0-2}{\sqrt{1^2+1^2}}\right|=\sqrt{2} \\ \mathrm{i}.\mathrm{e}.\left|\frac{-2}{\sqrt{2}}\right|=\sqrt{2} \\ \therefore \mathrm{Distance}\mathrm{of}A\mathrm{from}BC=3\sqrt{2}=\left|\frac{a+b-2}{\sqrt{1^2+1^2}}\right| \\ \mathrm{i}.\mathrm{e}.\left|a+b-2\right|=6 \\ \mathrm{i}.\mathrm{e}.a+b-2=6\mathrm{or}a+b-2=-6 \\ \mathrm{i}.\mathrm{e}.a+b-8=0\mathrm{or}a+b+4=0 \\ \mathrm{i}.\mathrm{e}.a+a-8=0\mathrm{or}a+a-4=0 \\ \mathrm{i}.\mathrm{e}.a=4\mathrm{or}a=-2 \\ \therefore \mathrm{vertex}\mathrm{is}\left(-2,-2\right) \end{gathered}$$

Hence, the correct answer is option C.