Question

Show that the point A(3,-1) B(5,-1) and C(3,-3) are the vertices of a right angled isosceles triangel

Answer 1

$\huge{\underline{\underline {\bf{\red{AnswEr:-}}}}}$

★ $\large\bold{\underline{\sf{\pink{GivEn:-}}}}$

• Coordinates of A :- (3,-1)
• Coordinates of B :- (5,-1)
• Coordinates of C :- (3,-3)

$\implies$ Using Distance formula to find all the sides:-

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

Now, AB :-

• = $\sqrt{( 3 - 5)^2 + ( -1 + 1 )^2}$

$\ \ \ \ \ \ \ \ \ \ \ = 2 units$

Now, BC :-

• = $\sqrt{( 5 - 3)^2 + ( -1 + 3 )^2}$

$\ \ \ \ \ \ \ \ \ \ \ = \sqrt{(4 + 4}$

$\ \ \ \ \ \ \ \ \ \ \ = 2 \sqrt{2}$

Now, CA :-

• $\sqrt{( 3- 3)^2 + ( -3 + 1 )^2}$

$\ \ \ \ \ \ \ \ \ \ \ = 2 units$

$\implies$ Using Pythagoras theorem to show that they are vertices of a right angle isosceles triangle.

$(BC)^2 = (AB)^2 + (CA)^2$

Now,

• $(BC)^2 = (2 \sqrt{2})^2$
• = 8

And,

• $(AB)^2 + (CA)^2 = 2^2 + 2^2$
• = 8

Therefore,

$(BC)^2 = (AB)^2 + (CA)^2$

★ $\large\bold{\underline{\sf{\red{Hence, \; these \; are \; the \; vertices \; of \; right \; angled \; isosceles \; triangle.}}}}$

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