Question

Show that the points P(3,3), (-3,-3) and O(0,0) are the vertices of an isosceles right angled triangle.

Answer 1

​HELLO DEAR,

Vertices of the triangle are p(3, 3), B(-3 , -3), C(0 , 0)

Distance between two points = $\sf{\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2}}$
PB = √[(-3 - 3)² + (-3 - 3)²] = √(36 + 36) = 6√2
BO = √[(0 - 3)² + (0 + 3)²] = √(9 + 9) = 3√2
PO = √[(0 - 3)² + (0 - 3)²] = √(9 + 9) = 3√2
HENCE, BO = PO

Therefore, PBO is an isosceles triangle.

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Dear student,

As we know that in an isosceles triangle, two sides are equal.

here three vertices of triangle are C ( 3 , 3) A ( 0 , 0) B (-3 ,-3).

Now, calculate the distance between AC ,AB and CA

we know that distance between two points (x1,y1) (x2, y2)

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

So,AC =$\sqrt{(0-3)^{2}+ (0-3)^{2}}$

AC = $\sqrt{9+9} \\ \\ \sqrt{18} \\ \\ 3\sqrt{2}$

Now calculate AB = $\sqrt{(0+3)^{2}+ (0+3)^{2}}$

AB = $\sqrt{9+9} \\ \\ \sqrt{18} \\ \\ 3\sqrt{2}$

Two sides AB and AC are equal.

So triangle formed by given vertices is isosceles.