Question

If $(3,2),(-3,2),(0,h)$ are vertices of an equilateral triangle and $h<0$, then $h=$

• A. $\sqrt{3}$
• B. $\sqrt{2}\pm \sqrt{27}$
• C. $2-\sqrt{27}$
• D. $2+\sqrt{27}$

Answer 1

The correct option is A $\sqrt{3}$

let the points be

$A(3,2)$

$B(3,-2)$

$C(0,h)$

$ThedistancebetweenAandB$

$=\surd (x2-x1)²+(y2-y1)²$

$=\surd (3-3)²+(-2-2)²$

$=\surd 0+16$

$=\surd \left(16\right)$

$=4$

$ThedistancebetweenBandC$

$=\surd (x2-x1)²+(y2-y1)²$

$=\surd (0-3)²+(-h-2)²$

$=\surd 9+h²+4+4h$

$=\surd 13+h²+4h$

The ,distance between $CandA$

$=\surd (x2-x1)²+(y2-y1)²$

$=\surd (0-3)²+(h-2)²$

$=\surd 9+h²+4-4h$

$=\surd 13+h²-4h$

as it is a equilateral triangle

$AB=BC=CA$

$=\surd 13+h²+4h=\surd 13+h²-4h=4$

we get

$h²+4h-3=0$

$h²-4h-3=0$

adding the equations we get

$2h²-6=0$

$h²-3=0$

$h²=3$

$h=\surd 3$

Hence, option $A$ is the correct answer.