Question

Let us prove that the three points A(3, 3), B (8,-2) and C(-2,-2) are the vertices of a right-angled isosceles triangle. Let us calculate the length of the hypotenuse of ∆ABC

Answer 1

Answer:

Yes, ΔABC is an isosceles right-angled triangle.

Step-by-step explanation:

If a right-angled triangle is isosceles, two sides except the hypotenuse are equal to each other. Using the distance formula, we'll try to prove that both the sides are equal.

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

If ΔABC is isoceles, AB = AC.

$\sf \Longrightarrow AB = AC$

$\sf \Longrightarrow \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} = \sqrt{(x_1-x_3)^2 + (y_1-y_3)^2}$

$\sf \Longrightarrow \sqrt{(3-8)^2 + (3+2)^2} = \sqrt{(3+2)^2 + (3+2)^2}$

Squaring on both sides we get,

$\sf \Longrightarrow \left(\sqrt{(3-8)^2 + (3+2)^2}\right)^{ 2} = \left(\sqrt{(3+2)^2 + (3+2)^2}\right)^2$

Squares and roots get cancelled.

$\sf \Longrightarrow (3-8)^2 + (3+2)^2 =(3+2)^2 + (3+2)^2$

$\sf \Longrightarrow (-5)^2 + \bcancel{(3+2)^2}=\bcancel{(3+2)^2} + (5)^2$

$\sf \Longrightarrow (-5)^2=(5)^2$

$\sf \Longrightarrow 25=25$

LHS = RHS.

∴ ABC is an isosceles triangle.

Alternative method:

Finding values of AB, BC and AC.

$\sf AB = \sqrt{(3-8)^2+(3+2)^2}\\ \\\sf AB = \sqrt{(-5)^2+(5)^2}\\ \\\sf AB = \sqrt{25+25}\\ \\\sf AB = \sqrt{50}\\ \\ \boxed{\sf AB = 5\sqrt{2}}$

$\sf AC = \sqrt{(3+2)^2+(3+2)^2}\\ \\\sf AC = \sqrt{(5)^2+(5)^2}\\ \\\sf AC = \sqrt{25+25}\\ \\\sf AC = \sqrt{50}\\ \\ \boxed{\sf AC = 5\sqrt{2}}$

$\sf BC = \sqrt{(8+2)^2+(-2+2)^2}\\ \\\sf BC = \sqrt{(10)^2+(0)^2}\\ \\\sf BC = \sqrt{100}\\ \\ \boxed{\sf BC = 10}$

Is ABC is a right-angled triangle, It'll follow the Pythagoras theorem.

$\boxed{\sf Pythagoras \ Theorem \longrightarrow Hypotenuse^2 = Base^2 + Altitude^2 \ }$

$\Longrightarrow \sf BC^2 = AC^2 + AB^2$

Substitute the values of AB, BC and AC above.

$\Longrightarrow \sf (10)^2 = (5\sqrt{2})^2 + (5\sqrt{2})^2\\ \\\Longrightarrow \sf 100 = (25 \times 2)+(25 \times 2)\\ \\\Longrightarrow \sf 100 = 50+50\\ \\\Longrightarrow \sf 100 = 100\\ \\\Longrightarrow \sf LHS = RHS$

∴ ABC is a right-angled triangle.

∴ Hypotenuse = BC = 10 units.

Answer 2

$\huge\bold\green{Answer}$

°•° Condition for right angle :-

→ Their 2 sides are equal

→ The two non-right angles must necessarily be acute

→ Pythagoras theorem can be used

So , firstly we used Distance Formula

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

So, as shown in picture ΔABC is isoceles, AB = AC.

=$\tt \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} = \sqrt{(x_1-x_3)^2 + (y_1-y_3)^2}$

=$\tt \sqrt{(3-8)^2 + (3+2)^2} = \sqrt{(3+2)^2 + (3+2)^2}$

Now , Squaring on both sides for cancellation of the roots :-

=$\tt{(3-8)^2 + (3+2)^2} = (3+2)^2 + (3+2)^2}$

=$\tt (-5)^2 + \bcancel{(3+2)^2}=\bcancel{(3+2)^2}$

= -5² = 5²

= 25 = 25 [RHS = LHS]

Hence RHS = LHS , So we can conclude that ABC is an isosceles triangle

As , we said in conditions above it can follows Pythagoras Theorem also. So, lets used it :-

$\tt\green{Hypotenuse^2 = Base^2 + Altitude^2 }$

= BC² = AC² + AB²

Niw by substituting the known values in formula :-

$\begin{lgathered} = \tt (10)^2 = (5\sqrt{2})^2 + (5\sqrt{2})^2\\ \\ = \tt 100 = (25 \times 2)+(25 \times 2)\\ \\ = \tt 100 = 50+50\\ \\= \tt 100 = 100\\ \end{lgathered}$

So, RHS is equals to LHS , hence we conclude that ABC is a right-angled triangle.

Hence the required Hypotenuse = BC = 10 units