Question

In the given figure, ∆ABC is an equilateral triangle of side 3 units. Find the coordinates of the other two vertices ?

Answer 1

Let the coordinates of C be (a, b). 

Since ∆ABC is an equilateral triangle so, AB = AC = BC = 3 units.

Now,

\[AC = \sqrt{\left( a - 2 \right)^2 + \left( b - 0 \right)^2}\]

\[\text{Squaring on both sides, we have}\]

\[ {AC}^2 = \left( a - 2 \right)^2 + \left( b \right)^2 \]

\[ \Rightarrow {AC}^2 = a^2 + 4 - 4a + b^2 . . . . . \left( 1 \right)\]

Since AB = 3 units, so the coordinates of B will be (5, 0) as the point B is 3 units away from A(2, 0) on the x-axis.

\[\therefore BC = \sqrt{\left( a - 5 \right)^2 + \left( b - 0 \right)^2}\]

\[\text{Squaring on both sides, we have}\]

\[ {BC}^2 = \left( a - 5 \right)^2 + \left( b \right)^2 \]

\[ \Rightarrow {BC}^2 = a^2 + 25 - 10a + b^2 . . . . . \left( 2 \right)\]

From (1) and (2), we have

\[a^2 + 4 - 4a + b^2 = a^2 + 25 - 10a + b^2 \]

\[ \Rightarrow 10a - 4a = 25 - 4\]

\[ \Rightarrow 6a = 21\]

\[ \Rightarrow a = \frac{21}{6} = \frac{7}{2}\]

Substituting `a=7/2` in (1), we have

\[\Rightarrow \left( 3 \right)^2 = \left( \frac{7}{2} - 2 \right)^2 + b^2 \]
\[ \Rightarrow 9 = \left( \frac{3}{2} \right)^2 + b^2 \]
\[ \Rightarrow b^2 = 9 - \frac{9}{4} = \frac{27}{4}\]
\[ \Rightarrow b = \pm \frac{3\sqrt{3}}{2}\]

Since C lies in the first quadrant, so \[b = \frac{3\sqrt{3}}{2}\]

. (In the first quadrant, x-coordinate and y-coordinate are both positive)

So, the coordinates of C are\[\left( \frac{7}{2}, \frac{3\sqrt{3}}{2} \right)\]

 

Thus, the coordinates of the other two vertices are B(5, 0) and C

\[\left( \frac{7}{2}, \frac{3\sqrt{3}}{2} \right)\]