Question

For an equilateral triangle, one side lies on $x+y=6$ and the $3^{rd}$ vertex is mirror image of the origin in the mirror $x+y=6$. Then the coordinates of other two vertices are

• A. $(3+\sqrt{3},3-\sqrt{3})$
• B. $(3+\sqrt{3},3+\sqrt{3})$
• C. $(3-\sqrt{3},3+\sqrt{3})$
• D. $(3-\sqrt{3},3-\sqrt{3})$

Answer 1

Answer: (A), (C)

The correct options are
A $(3+\sqrt{3},3-\sqrt{3})$
C $(3-\sqrt{3},3+\sqrt{3})$

​​​​Let $P(x,y)$ be the image of origin about $x+y=6$


The line perpendicular to $x+y=6$ and passes through $(0,0)$ will be $y=x$
So, intersection of $x=y$ and $x+y=6$ will give the midpoint of $QR$ i.e., $T$
$\Rightarrow T\equiv (3,3)$
Now, $OT=TP=3\sqrt{2}$
$\therefore QR\sin {60}^{\circ }=3\sqrt{2}(\because RP=QR=PQ)$
$\therefore QR=2\sqrt{6}$
Now, $x+y=6$
So any point on the line will be
$(3\pm r\cos 135°,3\pm r\sin 135°)=(3\mp \frac{r}{\sqrt{2}},3\pm \frac{r}{\sqrt{2}})$
where $r=TQ=\sqrt{6}$
So, from the above condition
$x=3\mp \sqrt{3}$ and $y=3\pm \sqrt{3}$