Question

Show that the points (3, 3), (9, 0) and (12, 21) are the vertices of a right-angled triangle.
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Answer 1

Answer:

Let A → (7, 9)

B → (3, -7)

C → (-3, 3)

be the vertices of a triangle.

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

.

Using the above distance formula,

AB

BC

CA

Now, (BC)2 + (CA)2

Now, (BC)2 + (CA)2and (AB)2

Now, (BC)2 + (CA)2and (AB)2

Answer 2

Question :-- Show that the points (3, 3), (9, 0) and (12, 21) are the vertices of a right-angled triangle.

Concept used :--

→ Distance Formula = √[(x2-x1)² + (y2-y1)²]

→ Pythagoras Theoram = Sum of square of two sides = square of third side ..

Solution :----

Let points of vertex A = (x1,y1) = (3,3)

Point of vertex B = (x2,y2) = (9,0)

Point of vertex C = (x3,y3) = (12,21)

Putting values now, in distance Formula we get,

→ Length of AB = √(9-3)² + (0-3)² = √36+9 = √45 units .

→ Length of BC = √(12-9)² + (21-0)² = √9+441 = √450 units .

→ Length of AC = √(12-3)² + (21-3)² = √81+324 = √405 units .

Now, by pythagoras theoram ,,

→ AB² + AC² = BC²

Putting values we get,

→ (√45)² + (√405)² = (√450)²

→ 45 + 405 = 450

→ 450 = 450 = Proved .

Hence, Points (3, 3), (9, 0) and (12, 21) are the vertices of a right-angled triangle.