Question

A^ 3,2h , B (0,1), C(0,3) Show that the points taken in order form an equilateral triangle in each case.

Answer 1

Hello,

Solution:

If given points are vertex of Equilateral triangle,then distance between any two are same.

Distance Formula:

A (x1,y1) B(x2, y2)

$AB = \sqrt{ {(x1 - x2) }^{2} + ( {y1 - y2)}^{2} }$

A (3,2) B (0,1)

$AB = \sqrt{( {3 - 0)}^{2}+ {(2 - 1)}^{2} } \\ \\ = \sqrt{9+ 1} \\ \\ = \sqrt{10}$

B(0,1) C(0,3)

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

A(3,2) C(0,3)

$AC = \sqrt{9+ 1} \\ = \sqrt{10}$

it is an isosceles Triangle not Equilateral triangle.

Hope it helps you.

Answer 2

Let the traingle  ABC  as  shown in figure,

Now,

We know that distance between two points  having  coordinates (x₁,y₁) and (x₂,y₂) is  

= $\sqrt{(x_1-x_2)^{2} +(y_1-y_2)^{2}}$

Using  this formula,

Distance between A(3,2) and  B(0,1)

$AB=\sqrt{(3-0)^{2} +(2-1)^{2}}\\\\AB=\sqrt{3^{2} +1^{2}}\\\\AB=\sqrt{9+1}\\\\AB=\sqrt{10}$

Distance between B(0,1) and  C(0,3)

$BC=\sqrt{(0-0)^{2} +(1-3)^{2}}\\\\BC=\sqrt{0 +2^{2}}\\\\BC=\sqrt{4}\\\\BC=2$

Distance between A(3,2) and  C(0,3)

$AC=\sqrt{(3-0)^{2} +(2-3)^{2}}\\\\AC=\sqrt{3^{2} +(-1)^{2}}\\\\AC=\sqrt{9+1}\\\\AC=\sqrt{10}$

AC=AB≠BC

It is an isosceles  Δ