Question

$(3,1),(-3,2)$ and $(0,2-\sqrt{3})$ are the vertices of __________ triangle of area ___________.

• A. an isosceles, $81$ sq. units
• B. a scalene, $\frac{-3+6\sqrt{3}}{2}$ sq. units
• C. an equilateral, $9\sqrt{3}$ sq. units
• D. a right angled, $81$ sq. units

Answer 1

Answer: (B)

a scalene, $\frac{-3+6\sqrt{3}}{2}$ sq. units

Distance between two points $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
Distance between the points $A(3,1)$ and $B(-3,2)=\sqrt{{(-3-3)}^2+{(2-1)}^2}=\sqrt{36+1}=\sqrt{37}$

Distance between the points $B(-3,2)$ and $C(0,2-\sqrt{3})=\sqrt{{(0+3)}^2+{(2-\sqrt{3}-2)}^2}=\sqrt{9+3}=\sqrt{12}$

Distance between the points $A(3,1)$ and $C(0,2-\sqrt{3})$

$=\sqrt{{(0-3)}^2+{(2-\sqrt{3}-1)}^2}=\sqrt{9+1+3-2\sqrt{3}}=\sqrt{13-2\sqrt{3}}$

Since the length of the sides between all vertices are different, they are the vertices of a scalene triangle.

Area of a triangle $=\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$
Area of $△ABC=\left|\frac{\left(3\right)(2-2+\sqrt{3})+(-3)(2-\sqrt{3}-1)+0(1-2)}{2}\right|$

$=\left|\frac{3\sqrt{3}-3+3\sqrt{3}}{2}\right|$

$=\frac{-3+6\sqrt{3}}{2}$ sq. units