9b) | Prove that the points (0,0),(0,3) and (3,0) are the vertices right angled isosceles triangle.
Answers
Answer:
Vertices of the triangle are A(−3,0), B(1,−3), C(4,1).
Distance between two points =
(
x
2
−x
1
)
2
+(y
2
−y
1
)
2
AB=
(1+3)
2
+(−3−0)
2
=5
BC=
(4−1)
2
+(1+3)
2
=5
AC=
(4+3)
2
+(1−0)
2
=5
2
AB=BC
Therefore, ΔABC is an isosceles triangle.
(AB)
2
+(BC)
2
=5
2
+5
2
=50
and (AC)
2
=(5
2
)
2
=50
∴(AB)
2
+(BC)
2
=(AC)
2
So, the triangle satisfies the Pythagoras theorem and hence it is a right angled triangle.
Step-by-step explanation:
Let (0,0) coordinate be O
Let (0,3) coordinate be P
Let (3,0) coordinate be Q
distance between O and P = 3 cm [graphically]
Distance between O and Q = 3 cm [graphically]
Now after joining P and Q , we get a triangle
it's two side,other than hypotenuse, are same .
And the x-axis and y-axis together give 90° to the triangle.
So points P, O,Q are the vertices of a right angled isosceles triangle