Question

show that the points (0,3),(0,1)and(2,3) are vertices of an equilateral triangle

Answer 1

Answer:

The given points form an isoceles right angled triangle

Step-by-step explanation:

Distance formula:

The distance between two points $(x_1,y_1)\:and\:(x_2,y_2)$ is

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

Let the given points be A(0,3), B(0,1) and C(2,3)

$AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\AB=\sqrt{(0-0)^2+(3-1)^2}\\\\AB= \sqrt{(0)^2+(2)^2}\\\\AB= \sqrt{4}=2$

$BC=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\BC=\sqrt{(0-2)^2+(1-3)^2}\\\\BC=\sqrt{(-2)^2+(-2)^2}\\\\BC=\sqrt{4+4}=\sqrt{8}$

$AC=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\AC=\sqrt{(0-2)^2+(3-3)^2}\\\\AC=\sqrt{(-2)^2+(0)^2}\\\\AC=\sqrt{4+0}=2$

AB=AC

Also,

$(AB)^2+(AC)^2\\\\=2^2+2^2\\\\=4+4\\\\=8\\\\=(BC)^2$

Therefore triangle ABC is an isoceles right angled triangle.

Answer 2

Answer:

given points are the vertices of an isosceles right angled triangle

Step-by-step explanation:

show that the points (0,3),(0,1)and(2,3) are vertices of an equilateral triangle

Question statement should be show that given points are the vertices of an isosceles right angled triangle

These points will be vertices of an equilateral triangle if distance between all the points are equal

a = (0,3)

b = (0,1)

c = (2,3)

as distance beytween two points (x1 , y1) & (x2 , y2)

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

so using this

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

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

bc = $\sqrt{(2-0)^2 + (3-1)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$

Hence ab = ac ≠ bc

so these are not vertices an equilateral triangle

ab² = 4

ac² = 4

bc² = 8

8 = 4 + 4

so bc² = ab² + ac² Hence right angles triangle

ab = ac so isosceles triangle

this is an isosceles right angled triangle

Hence given points are the vertices of an isosceles right angled triangle